## CHAPTER 3 Exponential and Logarithmic Functions

C H A P T E R 3 Exponential and Logarithmic Functions Section 3.1 Exponential Functions and Their Graphs . . . . . . . . . 265 Section 3.2 Logarith...
Author: Lindsey Hopkins
C H A P T E R 3 Exponential and Logarithmic Functions Section 3.1

Exponential Functions and Their Graphs . . . . . . . . . 265

Section 3.2

Logarithmic Functions and Their Graphs

Section 3.3

Properties of Logarithms . . . . . . . . . . . . . . . . . 281

Section 3.4

Exponential and Logarithmic Equations . . . . . . . . . 289

Section 3.5

Exponential and Logarithmic Models

Review Exercises

. . . . . . . . 273

. . . . . . . . . . 303

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 317

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 Practice Test

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333

C H A P T E R 3 Exponential and Logarithmic Functions Section 3.1

Exponential Functions and Their Graphs

You should know that a function of the form f x  a x, where a > 0, a  1, is called an exponential function with base a.

You should be able to graph exponential functions.

You should know formulas for compound interest.



(a) For n compoundings per year: A  P 1 

r n

. nt

(b) For continuous compoundings: A  Pert.

Vocabulary Check 1. algebraic



4. A  P 1 

2. transcendental r n



nt

3. natural exponential; natural

5. A  Pert

1. f 5.6  3.45.6  946.852

2. f x  2.3x  2.332  3.488

3. f    5  0.006

4. f x  23   23 

5. gx  50002x  500021.5

6. f x  2001.212x

5x

50.3

 0.544

 2001.212  24

 1767.767

 1.274  1025 7. f x  2x

9. f x  2x

8. f x  2x  1 rises to the right.

Increasing

Asymptote: y  1

Decreasing

Asymptote: y  0

Intercept: 0, 2

Asymptote: y  0

Intercept: 0, 1

Matches graph (c).

Intercept: 0, 1 Matches graph (a).

Matches graph (d). 10. f x  2x2 rises to the right. Asymptote: y  0 1 Intercept: 0, 4 

Matches graph (b).

11. f x  12 

x

y 5

x

2

1

0

1

2

f x

4

2

1

0.5

0.25

4 3 2

Asymptote: y  0

1 −3

−2

x

−1

1

2

3

−1

265

266

Chapter 3

12. f x  12 

x

Exponential and Logarithmic Functions 13. f x  6x

 2x

x

2

1

0

1

2

x

2

1

0

1

2

f x

0.25

0.5

1

2

4

f x

36

6

1

0.167

0.028

Asymptote: y  0

Asymptote: y  0

y

y 5

5

4

4

3

3

2 1 −3

−2

−1

x 1

2

−3

3

−2

−1

−1

x 1

2

3

−1

15. f x  2x1

14. f x  6x x

2

1

0

1

2

x

2

1

0

1

2

f x

0.028

0.167

1

6

36

f x

0.125

0.25

0.5

1

2

Asymptote: y  0

Asymptote: y  0 y

y

5

5

4

4

3

3

2

2

1 −3

−2

−1

1 x 1

2

−3

3

−2

x

−1

1

2

3

−1

−1

16. f x  4x3  3

y 7

x

1

0

1

2

3

f x

3.004

3.016

3.063

3.25

4

6 5 4

Asymptote: y  3

2 1 −3 −2 −1

17. f x  3x, gx  3x4 Because gx  f x  4, the graph of g can be obtained by shifting the graph of f four units to the right. 19. f x  2x, gx  5  2x Because gx  5  f x, the graph of g can be obtained by shifting the graph of f five units upward.

x 1

2

3

4

5

18. f x  4x, gx  4x  1 Because gx  f x  1, the graph of g can be obtained by shifting the graph of f one unit upward. 20. f x  10x, gx  10x3 Because gx  f x  3, the graph of g can be obtained by reflecting the graph of f in the y-axis and shifting f three units to the right. (Note: This is equivalent to shifting f three units to the left and then reflecting the graph in the y-axis.)

Section 3.1 21. f x  72  , gx   72 

x6

x

gx  f x  5, hence the graph of g can be obtained by reflecting the graph of f in the x-axis and shifting the resulting graph five units upward.

24. y  3x

2

25. f x  3x2  1

3

3

26. y  4x1  2 3

4

−6 −3

267

22. f x  0.3x, gx  0.3x  5

Because gx  f x  6, the graph of g can be obtained by reflecting the graph of f in the x-axis and y-axis and shifting f six units to the right. (Note: This is equivalent to shifting f six units to the left and then reflecting the graph in the x-axis and y-axis.) 23. y  2x

Exponential Functions and Their Graphs

−3

3

3

3 −1

−1

−1

5

−3

0

3 27. f 4   e34  0.472

28. f x  ex  e3.2  24.533

29. f 10  2e510  3.857  1022

30. f x  1.5e12x

31. f 6  5000e0.066  7166.647

32. f x  250e0.05x  250e0.0520  679.570

 1.5e120  1.956  1052 33. f x  e x

34. f x  ex

x

2

1

0

1

2

x

2

1

0

1

2

f x

0.135

0.368

1

2.718

7.389

f x

7.389

2.718

1

0.368

0.135

Asymptote: y  0

Asymptote: y  0

y

y

5

5

4

4

3

3

2

2

1 −3

−2

1 x

−1

1

2

−3

3

−1

−2

−1

x 1

2

3

−1

36. f x  2e0.5x

35. f x  3e x4 x

8

7

6

5

4

x

2

1

0

1

2

f x

0.055

0.149

0.406

1.104

3

f x

5.437

3.297

2

1.213

0.736

Asymptote: y  0

Asymptote: y  0

y

y 8

6

7

5

6 5

4

4

3

3

2

2

1

1 − 8 − 7 − 6 − 5 − 4 −3 −2 − 1

x 1

− 3 − 2 −1 −1

x 1

2

3

4

268

Chapter 3

Exponential and Logarithmic Functions

37. f x  2e x2  4

38. f x  2  ex5

x

2

1

0

1

2

x

0

2

4

5

6

f x

4.037

4.100

4.271

4.736

6

f x

2.007

2.050

2.368

3

4.718

Asymptote: y  4

Asymptote: y  2 y

y

−3 −2 −1

9 8 7 6 5

8

3 2 1

3

7 6 5 4

1 x

−1

1 2 3 4 5 6 7

3

4

5

6

7

41. st  2e0.12t 22

−4

−7

8

5 −10

−2

−1

42. st  3e0.2t

44. hx  ex2 4

4

− 16

17

−3

−2

−2

3

46.

3x1  33

4 0

0

3x1  27

23 0

43. gx  1  ex

20

2x3  16

47.

2x3  24

1 2x2  32

2x2  25

x13

x34

x  2  5

x2

x7

x  3

15 x1  125 15 x1  53 15 x1  15 3

49.

e3x2  e3

50.

2x  1  4

3x  1

2x  5

1 3

x  52

x  4 ex

2 3

 e2x

x 2  3  2x

52.

ex

2 6

 e5x

x 2  6  5x

x 2  2x  3  0

x 2  5x  6  0

x  3x  1  0

x  3x  2  0

x  3 or x  1

e2x1  e4

3x  2  3 x

x  1  3

51.

8

6

7

48.

2

40. y  1.085x

39. y  1.085x

45.

x 1

x  3 or x  2

Section 3.1

Exponential Functions and Their Graphs

53. P  \$2500, r  2.5%, t  10 years



Compounded n times per year: A  P 1 

r n



nt



 2500 1 

0.025 n



10n

Compounded continuously: A  Pert  2500e0.02510 n

1

2

4

12

365

Continuous Compounding

A

\$3200.21

\$3205.09

\$3207.57

\$3209.23

\$3210.04

\$3210.06

54. P  \$1000, r  4%, t  10 years



Compounded n times per year: A  1000 1 

0.04 n



10n

Compounded continuously: A  1000e0.0410

n

1

2

4

12

365

Continuous Compounding

A

\$1480.24

\$1485.95

\$1488.86

\$1490.83

\$1491.79

\$1491.82

55. P  \$2500, r  3%, t  20 years



Compounded n times per year: A  P 1 

r n



nt



 2500 1 

0.03 n



20n

Compounded continuously: A  Pert  2500e0.0320 n

1

2

4

12

365

Continuous Compounding

A

\$4515.28

\$4535.05

\$4545.11

\$4551.89

\$4555.18

\$4555.30

56. P  \$1000, r  6%, t  40 years



Compounded n times per year: A  1000 1 

0.06 n



40n

Compounded continuously: A  1000e0.0640

n

1

2

4

12

365

Continuous Compounding

A

\$10,285.72

\$10,640.89

\$10,828.46

\$10,957.45

\$11,021.00

\$11,023.18

57. A  Pert  12,000e0.04t t

10

20

30

40

50

A

\$17,901.90

\$26,706.49

\$39,841.40

\$59,436.39

\$88,668.67

58. A  Pert  12,000e0.06t t

10

20

30

40

50

A

\$21,865.43

\$39,841.40

\$72,595.77

\$132,278.12

\$241,026.44

269

270

Chapter 3

Exponential and Logarithmic Functions

59. A  Pert  12,000e0.065t t

10

20

30

40

50

A

\$22,986.49

\$44,031.56

\$84,344.25

\$161,564.86

\$309,484.08

60. A  Pert  12,000e0.035t t

10

20

30

40

50

A

\$17,028.81

\$24,165.03

\$34,291.81

\$48,662.40

\$69,055.23

61. A  25,000e0.087525

62. A  5000e0.07550

 \$222,822.57



64. p  5000 1  (a)

63. C10  23.951.0410  \$35.45

 \$212,605.41

4 4  e0.002x



65. Vt  100e4.6052t (a) V1  10,000.298 computers

1200

(b) V1.5  10,004.472 computers (c) V2  1,000,059.63 computers 0

2000 0

(b) When x  500:



p  5000 1 



4  \$421.12 4  e0.002500

(c) Since 600, 350.13 is on the graph in part (a), it appears that the greatest price that will still yield a demand of at least 600 units is about \$350. 67. Q  2512 

t1599

66. (a) P  152.26e0.0039t

(a) Q0  25 grams

Since the growth rate is negative, 0.0039  0.39%, the population is decreasing.

(b) Q1000  16.21 grams

(b) In 1998, t  8 and the population is given by P8  152.26e0.00398  147.58 million.

(c)

30

In 2000, t  10 and the population is given by P10  152.26e0.003910  146.44 million. 0

(c) In 2010, t  20 and the population is given by P20  152.26e0.003920  140.84 million. t5715

(a) When t  0: Q  1012 

05715

 101  10 grams (b) When t  2000: Q  102 

1 20005715

 7.85 grams

(c)

Q

Mass of 14C (in grams)

68. Q  1012 

5000 0

12 10 8 6 4 2 t 4000

8000

Time (in years)

Exponential Functions and Their Graphs

x

Sample Data

Model

0

12

12.5

25

44

44.5

50

81

81.82

75

96

96.19

100

99

99.3

271

100 1  7e0.069x

69. y  (a)

Section 3.1

(b)

110

0

120 0

70. (a)

y (d)

100  63.14%. 1  7e0.06936

2 100 100  when 3 1  7e0.069x x  38 masses.

(b) p  107,428e0.150h

P

Atmospheric pressure (in pascals)

(c) When x  36:

120,000

 107,428e0.1508

100,000

 32,357 pascals

80,000 60,000 40,000 20,000 h 5

10

15

20

25

Altitude (in km)

271,801 99,990 .

71. True. The line y  2 is a horizontal asymptote for the graph of f x  10x  2.

72. False, e 

73. f x  3x2

74. gx  22x6  22x  26

 3x32 1

 6422x

2

 6422x

3 

 3x

1  3x 9

 644x  hx

 hx Thus, f x  gx, but f x  hx. 75. f x  164x

e is an irrational number.

and

f x  164x

Thus, gx  hx but gx  f x.

76. f x  5x  3

 5x

 424x

 1622x

gx  53x  53

 42x

 1622x

hx  5x3   5x  53

 hx

Thus, none are equal.

2x

 14   4 

1 x2

 gx Thus, f x  gx  hx.

272

Chapter 3

Exponential and Logarithmic Functions

77. y  3x and y  4x y 3

y = 3x

y = 4x

x

2

1

0

1

2

3x

1 9

1 3

1

3

9

4x

1 16

1 4

1

4

16

2 1

−2

x

−1

1

(a) 4x < 3x when x < 0.

2

−1

(b) 4x > 3x when x > 0. (b) gx  x23x

78. (a) f x  x2ex

6

5

−2

−2

7

10 −2

−1

Decreasing:  , 0, 2, 

Decreasing: 1.44, 

Increasing: 0, 2

Increasing:  , 1.44

Relative maximum: 2, 4e2

Relative maximum: 1.44, 4.25

Relative minimum: 0, 0



79. f x  1 

0.5 x

 and gx  e x

0.5

80. The functions (c) 3x and (d) 2x are exponential.

(Horizontal line)

4

f g −3

3 0

As x → , f x → gx. As x →  , f x → gx.



81. x2  y2  25 y2

82. x  y  2

 25 

y  x  2 and y   x  2, x ≥ 2

y  ± 25  x2

83. f x 

2 9x

y 12

Vertical asymptote: x  9

9 6

Horizontal asymptote: y  0 x

11

10

f x

1

2



x2 y

x2

3

8

7

2

1

−18 − 15

−6 −3 −3 −6 −9

x 3

Section 3.2 84. f x  7  x

y

Domain:  , 7

Logarithmic Functions and Their Graphs

6 4

x

9

2

3

6

7

y

4

3

2

1

0

2 −4

−2

x 2

4

6

8

−2 −4 −6

Section 3.2

Logarithmic Functions and Their Graphs

You should know that a function of the form y  loga x, where a > 0, a  1, and x > 0, is called a logarithm of x to base a.

You should be able to convert from logarithmic form to exponential form and vice versa. y  loga x ⇔ ay  x

You should know the following properties of logarithms. (a) loga 1  0 since a0  1. (b) loga a  1 since a1  a. (c) loga ax  x since ax  ax . (d) aloga x  x Inverse Property (e) If loga x  loga y, then x  y.

You should know the definition of the natural logarithmic function. loge x  ln x, x > 0

You should know the properties of the natural logarithmic function. (a) ln 1  0 since e0  1. (b) ln e  1 since e1  e. (c) ln ex  x since ex  ex . (d) eln x  x

Inverse Property

(e) If ln x  ln y, then x  y. ■

You should be able to graph logarithmic functions.

Vocabulary Check 1. logarithmic

2. 10

4. aloga x  x

5. x  y

3. natural; e

1. log4 64  3 ⇒ 43  64

2. log3 81  4 ⇒ 34  81

1 3. log7 49  2 ⇒ 72  491

1 1 4. log 1000  3 ⇒ 103  1000

2 5. log32 4  5 ⇒ 3225  4

6. log16 8  34 ⇒ 1634  8

1 7. log36 6  2 ⇒ 36 12  6

8. log8 4  23 ⇒ 823  4

9. 53  125 ⇒ log5 125  3

10. 82  64 ⇒ log8 64  2

1 11. 8114  3 ⇒ log81 3  4

3 12. 932  27 ⇒ log9 27  2

273

274

Chapter 3

Exponential and Logarithmic Functions

1 1 13. 62  36 ⇒ log6 36  2

1 1 14. 43  64 ⇒ log4 64  3

15. 70  1 ⇒ log7 1  0

16. 103  0.001 ⇒ log10 0.001  3

17. f x  log2 x

18. f x  log16 x f 4  log16 4  12 since 1612  4

f 16  log2 16  4 since 24  16 19. f x  log7 x

20. f x  log x

f 1  log7 1  0 since 70  1

21. gx  loga x

f 10  log 10  1 since 101  10

ga2  loga a2  2 by the Inverse Property

22. gx  logb x g

b3

  logb

23. f x  log x b3

f

 3 since

4 5

24. f x  log x

  log   0.097

1 f 500   log 5001  2.699

4 5

b3  b3 f x  log x

25.

26. f x  log x

f 12.5  1.097

27. log3 34  4 since 34  34

f 75.25  1.877 29. log   1 since 1  .

28. log1.5 1

30. 9log9 15

Since 1.50  1, log1.5 1  0.

Since aloga x  x, 9log9 15  15.

31. f x  log4 x

32. gx  log6 x

y

Domain: x > 0 ⇒ The domain is 0, .

Domain: 0, 

2

x-intercept: 1, 0

x-intercept: 1, 0

1

y

Vertical asymptote: x  0

2

y  log4 x ⇒ 4 y  x

1

Vertical asymptote: x  0 y  log6 x ⇒ 6 y  x

1 4

1

4

2

f x

1

0

1

1 2

−1

1

2

3

−1 −2

33. y  log3 x  2 4

x 2 −2

2  log3 x

−4

32  x

−2

x

1 6

1

6

6

y

1

0

1 2

1

The domain is 3, .

2

log3 x  2  0

4

6

8

10

12

y

x-intercept:

6

log4x  3  0

4 2

40  x  3

−6

9x The x-intercept is 9, 0.

x 2

1x3

−2

4x

−4

4

Vertical asymptote: x  0

The x-intercept is 4, 0.

y  log3 x  2

Vertical asymptote: x  3  0 ⇒ x  3

log3 x  2  y ⇒

32y

x

3

Domain: x  3 > 0 ⇒ x > 3

6

x-intercept:

2

34. hx  log4x  3

y

Domain: 0, 

1 −1

x

x

x

−1

y  log4x  3 ⇒ 4 y  3  x

x

27

9

3

1

1 3

x

34

1

4

7

19

y

1

0

1

2

3

y

1

0

1

2

6

8

10

Section 3.2 35. f x  log6x  2

Logarithmic Functions and Their Graphs

36. y  log5x  1  4

Domain: x  2 > 0 ⇒ x > 2

Domain: x  1 > 0 ⇒ x > 1

The domain is 2, .

The domain is 1, .

y

x-intercept: 0  log6x  2

4

x-intercept:

2

log5x  1  4  0

0  log6x  2

6 −2

1x2

−4

The x-intercept is 1, 0.

1

0

1

37. y  log

135 36 2

5

Domain:

x

x > 0 ⇒ x > 0 5

x-intercept:

x1

626 625

x

1 x 2

3

4

x

1.00032

1.0016

1.008

1.04

1.2

y

1

0

1

2

3

1

2

3

4

5

6

7

y

0.70

0.40

0.22

0.10

0

0.08

0.15

y 4

5  0 x

2 x 4

x  100 5

6

8

−2

x 1 ⇒ x5 5

−4

The x-intercept is 5, 0. Vertical asymptote:

1 625

x

The domain is 0, .

log

2

y  log5x  1  4 ⇒ 5y4  1  x

6y  2  x

f x

3

Vertical asymptote: x  1  0 ⇒ x  1

y  log6x  2

1

4

626

y  log6x  2

4

5

The x-intercept is 625, 0.

Vertical asymptote: x  2  0 ⇒ x  2

x

6

54  x  1

1  x

156

y

log5x  1  4

x

60  x  2



275

x 0 ⇒ x0 5

The vertical asymptote is the y-axis. 38. y  logx Domain: x > 0 ⇒ x < 0 The domain is  , 0.

x

1  100

 10

1

1

10

y

2

1

0

1

x-intercept: logx  0

y

100  x

2

1  x

1

The x-intercept is 1, 0.

−3

−2

x

−1

1

Vertical asymptote: x  0

−1

y  logx ⇒ 10y  x

−2

5

6

276

Chapter 3

Exponential and Logarithmic Functions

39. f x  log3 x  2

40. f x  log3 x

Asymptote: x  0

Asymptote: x  0

Point on graph: 1, 2

Point on graph: 1, 0

Matches graph (c).

Matches graph (f).

The graph of f x is obtained by shifting the graph of gx upward two units.

f x reflects gx in the x-axis.

41. f x  log3x  2

42. f x  log3x  1

Asymptote: x  2

Asymptote: x  1

Point on graph: 1, 0

Point on graph: 2, 0

Matches graph (d).

Matches graph (e).

The graph of f x is obtained by reflecting the graph of gx about the x-axis and shifting the graph two units to the left.

f x shifts gx one unit to the right.

43. f x  log31  x  log3 x  1

44. f x  log3x

Asymptote: x  1

Asymptote: x  0

Point on graph: 0, 0

Point on graph: 1, 0

Matches graph (b).

Matches graph (a).

The graph of f x is obtained by reflecting the graph of gx about the y-axis and shifting the graph one unit to the right.

f x reflects gx in the x-axis then reflects that graph in the y-axis.

45. ln 12  0.693 . . . ⇒ e0.693 . . .  12

46. ln 25  0.916 . . . ⇒ e0.916 . . .  25

47. ln 4  1.386 . . . ⇒ e1.386 . . .  4

48. ln 10  2.302 . . . ⇒ e2.302 . . .  10

49. ln 250  5.521 . . . ⇒ e5.521 . . .  250

50. ln 679  6.520 . . . ⇒ e6.520 .

51. ln 1  0 ⇒ e0  1

52. ln e  1 ⇒ e1  e

53. e3  20.0855 . . . ⇒ ln 20.0855 . . .  3

54. e2  7.3890 . . . ⇒ ln 7.3890 . . .  2

55. e12 1.6487 . . . ⇒ ln 1.6487 . . .  12

1 56. e13  1.3956 . . . ⇒ ln 1.3956 . . .  3

57. e0.5  0.6065 . . . ⇒ ln 0.6065 . . .  0.5

58. e4.1  0.0165 . . . ⇒ ln 0.0165 . . .  4.1

59. ex  4 ⇒ ln 4  x

60. e2x  3 ⇒ ln 3  2x

61. f x  ln x

62. f x  3 ln x

f 18.42  ln 18.42  2.913 63. gx  2 ln x g0.75  2 ln 0.75  0.575

f 0.32  3 ln 0.32  3.418 64. gx  ln x g12   ln 12  0.693

. .

 679

Section 3.2 65. gx  ln x

Logarithmic Functions and Their Graphs

66. gx  ln x

ge3  ln e3  3 by the Inverse Property

ge2  ln e2  2

67. gx  ln x

68. gx  ln x

ge23  ln e23 

 23

ge52  ln e52   52

by the Inverse Property

69. f x  lnx  1

70. hx  lnx  1

Domain: x  1 > 0 ⇒ x > 1

Domain: x  1 > 0 ⇒ x > 1

The domain is 1, .

The domain is 1, .

y 3

x-intercept:

2

0  lnx  1

1

e0  x  1

−1

2x

6

lnx  1  0

4

2

3

4

5

−1

2

3

4

f x

0.69

0

0.69

1.10

2

4

8

The x-intercept is 0, 0. Vertical asymptote: x  1  0 ⇒ x  1 y  lnx  1 ⇒ ey  1  x x

0.39

0

1.72

6.39

19.09

y

 12

0

1

2

3

72. f x  ln3  x

71. gx  lnx Domain: x > 0 ⇒ x < 0

Domain: 3  x > 0 ⇒ x < 3

y

The domain is  , 0.

2

The domain is  , 3.

x-intercept:

1

x-intercept:

0  lnx

−3

−2

y 3 2

ln3  x  0

x

−1

1

e0  3  x

e0  x 1  x

−2

−1

13x

−2

The x-intercept is 1, 0.

x 1

−3

The x-intercept is 2, 0.

x

0.5

1

2

3

Vertical asymptote: 3  x  0 ⇒ x  3

gx

0.69

0

0.69

1.10

y  ln3  x ⇒ 3  ey  x

74. f x  logx  1

73. y1  logx  1

x

2.95

2.86

2.63

2

0.28

y

3

2

1

0

1

75. y1  lnx  1 3

2

2

5

2

−1 −2

2x

Vertical asymptote: x  0 ⇒ x  0

−2

6

0x

Vertical asymptote: x  1  0 ⇒ x  1 1.5

x

−2

1x1

−3

x

2

e0  x  1

x 1

y

x-intercept:

−2

The x-intercept is 2, 0.

−1

277

−1

5

−2

0

−3

9

4

278

Chapter 3

Exponential and Logarithmic Functions

76. f x  lnx  2

78. f x  3 ln x  1

77. y  ln x  2 5

3

−4

4

−5

5 0

10

9

−1

−3

−6

80. log2x  3  log2 9

79. log2x  1  log2 4

x39

x14

x  12

x3 81. log2x  1  log 15

82. log5x  3  log 12

2x  1  15

5x  3  12

x7

5x  9 9

x5 83. lnx  2  ln 6

84. lnx  4  ln 2

x26

x42

x4

x6

85. lnx 2  2  ln 23

lnx 2  x  ln 6

86.

x 2  2  23

x2  x  6

x 2  25

x2  x  6  0

x  ±5

x  3x  2  0 x  2 or x  3

87. t  12.542 ln

x  x1000, x > 1000

(a) When x  \$1100.65:



88. t  (a)



1100.65 t  12.542 ln  30 years 1100.65  1000 When x  \$1254.68: t  12.542 ln

(b) Total amounts: 1100.651230  \$396,234.00 (c) Interest charges: 396,234  150,000  \$246,234 301,123.20  150,000  \$151,123.20 (d) The vertical asymptote is x  1000. The closer the payment is to \$1000 per month, the longer the length of the mortgage will be. Also, the monthly payment must be greater than \$1000.

K

1

2

4

6

8

10

12

t

0

7.3

14.6

18.9

21.9

24.2

26.2

The number of years required to multiply the original investment by K increases with K. However, the larger the value of K, the fewer the years required to increase the value of the investment by an additional multiple of the original investment.

1254.68  20 years 1254.68  1000 

1254.681220  \$301,123.20

ln K 0.095

(b)

t 25 20 15 10 5 K 2

4

6

8

10

12

Section 3.2

89. f t  80  17 logt  1, 0 ≤ t ≤ 12 (a)

Logarithmic Functions and Their Graphs

90.   10 log

10I  12

100

(a)   10 log

0

101   10 log10

(b)   10 log

12

279

0

  120 decibels

12

12 2

1010   10 log10

  100 decibels

10

12

(c) No, the difference is due to the logarithmic relationship between intensity and number of decibels.

(b) f 0  80  17 log 1  80.0 (c) f 4  80  17 log 5  68.1 (d) f 10  80  17 log 11  62.3 91. False. Reflecting gx about the line y  x will determine the graph of f x. 93. f x  3x, gx  log3 x

94. f x  5x, gx  log5 x

95 . f x  ex, gx  ln x

y

y 2

y

2

f

2

f

1

−2

92. True, log3 27  3 ⇒ 33  27.

g

1

g x

−1

f

1

1

−2

2

−1

1

−2

2

−1

−2

−2

−2

f and g are inverses. Their graphs are reflected about the line y  x.

40

g

The natural log function grows at a slower rate than the square root function.

2

f 1

f 0

x 1

4 x (b) f x  ln x, gx  

2

−1

15

g

The natural log function grows at a slower rate than the fourth root function.

−2

f

0

f and g are inverses. Their graphs are reflected about the line y  x.

(a)

1000 0

g −1

20,000 0

ln x x

x

1

5

10

102

104

106

f x

0

0.322

0.230

0.046

0.00092

0.0000138

(b) As x → , f  x  → 0. (c)

0.5

0

100 0

2

f and g are inverses. Their graphs are reflected about the line y  x.

97. (a) f x  ln x, gx  x

y

98. f x 

1

−1

96. f x  10x, gx  log10 x

−2

x

−1

−1

f and g are inverses. Their graphs are reflected about the line y  x.

g

x

280

Chapter 3

Exponential and Logarithmic Functions (b) True. y  loga x

99. (a) False. If y were an exponential function of x, then y  ax, but a1  a, not 0. Because one point is 1, 0, y is not an exponential function of x.

For a  2, y  log2 x. x  1, log2 1  0

(c) True. x  ay

x  2, log2 2  1

For a  2, x  2y.

x  8, log2 8  3

y  0, 20  1

(d) False. If y were a linear function of x, the slope between 1, 0 and 2, 1 and the slope between 2, 1 and 8, 3 would be the same. However,

y  1, 21  2 y  3, 23  8

m1 

10 31 2 1  1 and m2    . 21 82 6 3

Therefore, y is not a linear function of x. 100. y  loga x ⇒ ay  x, so, for example, if a  2, there is no value of y for which 2y  4. If a  1, then every power of a is equal to 1, so x could only be 1. So, loga x is defined only for 0 < a < 1 and a > 1.

101. f x  ln x (a)

102. (a) hx  lnx2  1 (b) Increasing on 1,  Decreasing on 0, 1

4

−1

8

(b) Increasing on 0,  Decreasing on  , 0

8

(c) Relative minimum: 0, 0

(c) Relative minimum: 1, 0

−9

−2

9

−4

For Exercises 103–108, use f x  3x  2 and g x  x3  1. 103.  f  g2  f 2  g2

104. f x  gx  3x  2  x3  1

 32  2  23  1

 3x  2  x3  1

87

 3x  x3  3

 15

Therefore,

 f  g1  31  13  3  3  1  3  1.

105.  fg6  f 6g6  36  2 63  1  20215

106.

f x 3x  2  3 gx x 1 f 3 02 Therefore, 0  3  2. g 0 1



 4300 107.  f g7  f g7

108. g f (x  g f x  g3x  2  3x  23  1

 f 73  1

Therefore,

 f 342

g f 3  3 3  23  1

 3342  2  1028

 73  1  344.

Section 3.3

Section 3.3 ■

Properties of Logarithms

You should know the following properties of logarithms. logb x log10 x ln x loga x  (a) loga x  loga x  log10 a ln a logb a (b) logauv  loga u  loga v (c) loga

 v   log u

a

lnuv  ln u  ln v

u  loga v

ln

(d) loga un  n loga u ■

Properties of Logarithms

 v   ln u  ln v u

ln un  n ln u

You should be able to rewrite logarithmic expressions using these properties.

Vocabulary Check log x ln x  log a ln a

1. change-of-base

2.

3. logauv  loga u  loga v This is the Product Property. Matches (c).

4. ln un  n ln u This is the Power Property. Matches (a).

u  loga u  loga v v This is the Quotient Property. Matches (b).

5. loga

1. (a) log5 x  (b) log5 x 

log x log 5 ln x ln 5

4. (a) log13 x  (b) log13 x 

7. (a) log2.6 x  (b) log2.6 x 

10. log7 4 

log x log13 ln x ln13 log x log 2.6 ln x ln 2.6

2. (a) log3 x  (b) log3 x 

5. (a) logx (b) logx

14. log20 0.125 

16. log3 0.015 

log 0.125 ln 0.125   0.694 log 20 ln 20

log 0.015 ln 0.015   3.823 log 3 ln 3

(b) log15 x 

log310 3  log x 10

6. (a) logx

3 ln310  10 ln x

(b) log7.1 x 

log 5 ln 5  1.161  log14 ln14

3. (a) log15 x 

ln x ln 3

8. (a) log7.1 x 

log 4 ln 4   0.712 log 7 ln 7

12. log14 5 

log x log 3

(b) logx

log x log 7.1

log x log15 ln x ln15

3 log34  4 log x 3 ln34  4 ln x

9. log3 7 

log 7 ln 7   1.771 log 3 ln 3

ln x ln 7.1

11. log12 4 

log 4 ln 4   2.000 log12 ln12

13. log90.4 

log 0.4 ln 0.4   0.417 log 9 ln 9

15. log15 1250 

17. log4 8 

log 1250 ln 1250   2.633 log 15 ln 15

log2 8 log2 23 3   log2 4 log2 22 2

281

282

Chapter 3

Exponential and Logarithmic Functions

18. log242  34  log2 42  log2 34

1 1 19. log5 250  log5125

 12 

 2 log2 4  4 log2 3

1  log5 125  log5 12

 2 log2

 log5

22

 4 log2 3

53

 log5

9 3 20. log 300  log 100

 log 3  log 100 21

 log 3  log 102

 3  log5 2

 4 log2 2  4 log2 3

 log 3  2 log 10

 4  4 log2 3

 log 3  2

21. ln5e6  ln 5  ln e6

22. ln

6  ln 6  ln e2 e2

 ln 5  6

23. log3 9  2 log3 3  2

 ln 6  2 ln e

 6  ln 5

 ln 6  2

1 24. log5 125  log5 53  3 log5 5  31  3

4 8  1 log 23  3 log 2  3 1  3 25. log2  4 2 4 2 4 4

3 6  log 613  1 log 6  1 1  1 26. log6  6 3 6 3 3

27. log4 161.2  1.2log4 16  1.2 log4 42  1.22  2.4

28. log3 810.2  0.2 log3 81

29. log39 is undefined. 9 is not in the domain of log3 x.

 0.2 log3 3

4

 0.24  0.8 30. log216 is undefined because 16 is not in the domain of log2 x.

31. ln e4.5  4.5

32. 3 ln e4  34 ln e  121  12

33. ln

1 e

4 e3  ln e34 34. ln 

 ln 1  lne 0



1 ln e 2



1 2

36. 2 ln e6  ln e5  ln e12  ln e5  ln

3 ln e 4

3  1 4

1  0  1 2 

35. ln e2  ln e5  2  5  7

e12 e5

3 4

37. log5 75  log5 3  log5

75 3

 log5 25

 ln e7

 log5 52

7

 2 log5 5 2

38. log4 2  log4 32  log4 412  log4 452 1 2

5 2

 log4 4  log4 4  121  521 3

39. log4 5x  log4 5  log4 x

Section 3.3

40. log3 10z  log3 10  log3 z

43. log5

5  log5 5  log5 x x

Properties of Logarithms

283

y  log y  log 2 2

41. log8 x4  4 log8 x

42. log

44. log6 z3  3 log6 z

45. lnz  ln z12 

47. ln xyz2  ln x  ln y  ln z2

48. log 4x2y  log 4  log x2  log y

1 ln z 2

 1  log5 x 3 t  ln t13  1 ln t 46. ln  3

 ln x  ln y  2 ln z 49. ln zz  12  ln z  lnz  12

50. ln

 ln z  2 lnz  1, z > 1



 log 4  2 log x  log y

x2  1  lnx2  1  ln x3 x3  lnx  1x  1  ln x3



 lnx  1  lnx  1  3 ln x

51. log2

a  1

9

 log2a  1  log2 9 

6

52. ln

x2  1

 ln 6  lnx2  112

1 log2a  1  log2 32 2

 ln 6 

1  log2a  1  2 log2 3, a > 1 2

xy  31 ln yx

xy

2

3

53. ln

 ln 6  lnx2  1

54. ln

3

 ln

1  ln x  ln y 3 1 1  ln x  ln y 3 3

y  x2

3

12

1 lnx2  1 2



 

1 x2 ln 3 2 y

1  ln x2  ln y3 2 1  2 ln x  3 ln y 2  ln x 

x z y   ln x 4

55. ln

4y

5

 ln z5

56. log2

1 ln y  5 ln z 2

yxz   log x  log y z 2 3

5

2

5

2 3

 log2 x y4  log2 z4



2

57. log5

z4

 log2 x  log2 y4  log2 z4

 ln x4  ln y  ln z5  4 ln x 

x y4

3 ln y 2

58. log

1 log2 x  4 log2 y  4 log2 z 2

xy4  log xy4  log z5 z5

 log5 x2  log5 y2  log5 z3

 log x  log y4  log z5

 2 log5 x  2 log5 y  3 log5 z

 log x  4 log y  5 log z

4 3 2 59. ln  x x  3  14 ln x3x2  3



1 3 4 ln x

 lnx2  3

60. lnx2x  2  lnx2x  2 12  lnxx  212

 14 3 ln x  lnx2  3

 ln x  lnx  212

 34 ln x  14 lnx2  3

 ln x  12 lnx  2

284

Chapter 3

Exponential and Logarithmic Functions

61. ln x  ln 3  ln 3x

64. log5 8  log5 t  log5

67.

8 t

62. ln y  ln t  ln yt  ln ty

63. log4 z  log4 y  log4

65. 2 log2x  4  log2x  42

66.

1 4 5x log3 5x  log35x14  log3  4

 ln

1 16x 4

70. 2 ln 8  5 lnz  4  ln 82  lnz  45  ln 64  lnz  45

x x  13

 ln 64z  45

71. log x  2 log y  3 log z  log x  log y2  log z3  log

2 log7z  2  log7z  223 3

68. 4 log6 2x  log62x4  log6

69. ln x  3 lnx  1  ln x  lnx  13

72. 3 log3 x  4 log3 y  4 log3 z  log3 x3  log3 y4  log3 z4

x xz3  log z3  log 2 y2 y

 log3 x3y4  log3 z4  log3

73. ln x  4lnx  2  lnx  2  ln x  4 lnx  2x  2  ln x  4 lnx2  4  ln x  lnx2  44  ln

x x2  44

74. 4ln z  lnz  5  2 lnz  5  4ln zz  5  lnz  52  lnzz  5 4  lnz  52  ln

75.

z y

z4z  54 z  52

1 1 2 lnx  3  ln x  lnx2  1  lnx  32  ln x  lnx2  1 3 3 1  ln xx  32  lnx2  1 3 

1 xx  32 ln 2 3 x 1

xxx  31

 ln

2

3

2

76. 23 ln x  lnx  1  lnx  1  2ln x3  lnx  1  lnx  1  2ln x3  lnx  1  lnx  1  2ln x3  lnx  1x  1  2 ln  ln

x2

x

2

x3 1

x3 1



2

x3y4 z4

Section 3.3

77.

Properties of Logarithms

1 1 log8 y  2 log 8 y  4  log 8 y  1  log 8 y  log 8 y  42  log 8 y  1 3 3 

1 log 8 y y  42  log 8 y  1 3

3  log 8  y  y  42  log 8 y  1

 log 8



3  y  y  42

y1



78. 12log4x  1  2 log4x  1  6 log4 x  12log4x  1  log4x  12  log4 x6  12log4x  1x  12  log4 x6  log4x  1x  1  log4 x6  log4x6x  1x  1

79. log2

32 log2 32  log2 32  log2 4  4 log2 4

The second and third expressions are equal by Property 2.

80. log770 

1 1 log7 70  log7 7  log7 10 2 2

81.   10 log

10I  12

1  1  log7 10 2

 10log I  log 1012

1 1   log7 10 2 2

 120  10 log I



 10log I  12

When I  106 :

1  log7 10 by Property 1 and Property 3 2

  120  10 log 106  120  106  60 decibels

82.   10 log

10I 

83.   120  10 log2I 

12

Difference  10 log

5

7

3.1610 10   10 log1.2610 10  12

12

 10log3.16  107  log1.26  105  10  3.16 1.26  10 

 10 log

7 5

 10log2.5079  102  10log250.79  24 dB

 120  10log 2  log I   120  10 log I   10 log 2 With both stereos playing, the music is 10 log 2  3 decibels louder.

285

286

Chapter 3

Exponential and Logarithmic Functions

84. f t  90  15 logt  1, 0 ≤ t ≤ 12 (a) f t  90  logt  115

(f) The average score will be 75 when t  9 months. See graph in (e).

(b) f 0  90

(g)

(c) f 4  90  15  log4  1  79.5

15  15 logt  1

(d) f 12  90  15  log12  1  73.3 (e)

75  90  15 logt  1 1  logt  1

95

101  t  1 t  9 months 0

12 70

85. By using the regression feature on a graphing calculator we obtain y  256.24  20.8 ln x. 86. (a)

(c)

80

0

30 0

(b) T  21  54.40.964 t T  54.40.964 t  21 See graph in (a). (d)

1  0.0012t  0.016 T  21 T

1  21 0.0012t  0.016

t (in minutes)

T C

T  21 C

lnT  21

1T  21

0

78

57

4.043

0.0175

5

66

45

3.807

0.0222

10

57.5

36.5

3.597

0.0274

15

51.2

30.2

3.408

0.0331

20

46.3

25.3

3.231

0.0395

25

42.5

21.5

3.068

0.0465

30

39.6

18.6

2.923

0.0538

5 0.07

80

0 0

30 0

0

30

30 0

0

(e) Since the scatter plot of the original data is so nicely exponential, there is no need to do the transformations unless one desires to deal with smaller numbers. The transformations did not make the problem simpler.

lnT  21  0.037t  4 T  e0.037t4  21 This graph is identical to T in (b).

Taking logs of temperatures led to a linear scatter plot because the log function increases very slowly as the x-values increase. Taking the reciprocals of the temperatures led to a linear scatter plot because of the asymptotic nature of the reciprocal function. 87. f x  ln x False, f 0  0 since 0 is not in the domain of f x. f 1  ln 1  0

88. f ax  f a  f x, a > 0, x > 0 True, because f ax  ln ax  ln a  ln x  f a  f x.

Section 3.3

89. False. f x  f 2  ln x  ln 2  ln

x  lnx  2 2

90. f x 

Properties of Logarithms

1 f x; false 2

f x  ln x can’t be simplified further.

f x   lnx  ln x12 

1 1 ln x  f  x  2 2

92. If f x < 0, then 0 < x < 1.

91. False. f u  2f v ⇒ ln u  2 ln v ⇒ ln u 

ln v2

⇒ u

 v2

True

93. Let x  logb u and y  logb v, then bx  u and by  v.

94. Let x  logb u, then u  bx and un  bnx. logb un  logb bnx  nx  n logb u

u bx  y  bxy v b Then logbuv  logbb xy  x  y  logb u  logb v.

95. f x  log2 x 

ln x log x  log 2 ln 2

96. f x  log4 x 

97. f x  log12 x 

2

3

−3

log x ln x  log 4 ln 4

6

−1

3

5

−3

−2

−3

log x ln x  log12 ln12

6

−3

99. f x  log11.8 x

98. f x  log14 x 

log x ln x  log14 ln14



log x ln x  log 11.8 ln 11.8



5

−1

−2

x ln x 101. f x  ln , gx  , hx  ln x  ln 2 2 ln 2 f x  hx by Property 2

log x ln x  log 12.4 ln 12.4

2

2

2

−1

100. f x  log12.4 x

−1

5

−2

5

−2

y 2 1

g

f=h x

1 −1 −2

2

3

4

287

288

Chapter 3

Exponential and Logarithmic Functions

102. ln 2  0.6931, ln 3  1.0986, ln 5  1.6094 ln 2  0.6931 ln 3  1.0986 ln 4  ln2

 2  ln 2  ln 2  0.6931  0.6931  1.3862

ln 5  1.6094 ln 6  ln2

 3  ln 2  ln 3  0.6931  1.0986  1.7917

ln 8  ln 23  3 ln 2  30.6931  2.0793 ln 9  ln 32  2 ln 3  21.0986  2.1972

 2  ln 5  ln 2  1.6094  0.6931  2.3025 ln 12  ln22  3  ln 22  ln 3  2 ln 2  ln 3  20.6931  1.0986  2.4848 ln 15  ln5  3  ln 5  ln 3  1.6094  1.0986  2.7080 ln 10  ln5

ln 16  ln 24  4 ln 2  40.6931  2.7724

 2  ln 32  ln 2  2 ln 3  ln 2  21.0986  0.6931  2.8903 ln 20  ln5  22  ln 5  ln 22  ln 5  2 ln 2  1.6094  20.6931  2.9956

ln 18  ln32

103.

24xy2 24xx3 3x4   ,x0 16x3y 16yy2 2y3

105. 18x3y4318x3y43 

107.

18x3y43  1 if x  0, y  0. 18x3y43

3x2  2x  1  0

104.

2x2

3



2x  3y

2

3



106. xyx1  y11 

108.

3y3 27y3  2 3 2x  8x 6

xy x1  y1



xy 1x  1y



xy2 xy   y  xxy x  y

4x2  5x  1  0

4x  1x  1  0

3x  1x  1  0 3x  1  0 ⇒ x 

 3y 

4x  1  0 ⇒ x  14

1 3

x10 ⇒ x1

x  1  0 ⇒ x  1

The zeros are x  14, 1. 2 x  3x  1 4

109.

5 2x  x1 3

110.

3x  1x  24

53  2xx  1

3x  x  8  0

15  2x2  2x

2

x

1 ± 12  438 23

1 ± 97  6

0  2x2  2x  15  2 ± 22  4215 x 22 2 ± 124 x 4 1 ± 31 x 2 The zeros are

1 ± 31 . 2

Section 3.4

Section 3.4 ■

Exponential and Logarithmic Equations

To solve an exponential equation, isolate the exponential expression, then take the logarithm of both sides. Then solve for the variable. 1. loga ax  x

2. ln ex  x

To solve a logarithmic equation, rewrite it in exponential form. Then solve for the variable. 1. aloga x  x

Exponential and Logarithmic Equations

2. eln x  x

If a > 0 and a  1 we have the following: 1. loga x  loga y ⇔ x  y 2. ax  ay ⇔ x  y

Check for extraneous solutions.

Vocabulary Check 2. (a) x  y (c) x

1. solve

1. 42x7  64 425 7

3. extraneous

2. 23x1  32

x5

(a)

(b) x  y (d) x



231 1

 64

Yes, x  5 is a solution. x2

(b)

1 64

No, x  2 is not a solution.

x  2  e25 

No, x  2  (b)

No, x  2 is not a solution. 4. 2e5x2  12

2e25 2

3e

25 3ee

e25

1 x  2  ln 6 5

(a)

 75

2e5152ln 6 2  2e2ln 62

is not a solution.

 2eln 6  2  6  12

x  2  ln 25

1 Yes, x  2  ln 6 is a solution. 5

3e2ln 25 2  3eln 25  325  75 Yes, x  2  ln 25 is a solution. (c)

x2 232 1  27  128

 64

3. 3ex2  75

 22  14

No, x  1 is not a solution. (b)

422 7  43 

(a)

x  1

(a)

43

x  1.219

x

(b)

3e1.2192  3e3.219  75

ln 6 5 ln 2

2e5[ln 65 ln 2 2  2eln 6ln 2 2

Yes, x  1.219 is a solution.

 2e2.5852  2  97.9995  195.999 No, x 

ln 6 is not a solution. 5 ln 2 x  0.0416

(c)

50.0416 2

2e

 2e1.792  26.00144  12

Yes, x  0.0416 is an approximate solution.

289

290

Chapter 3

Exponential and Logarithmic Functions

5. log43x  3 ⇒ 3x  43 ⇒ 3x  64

6. log2x  3  10

x  21.333

(a)

x  1021

(a)

321.333  64

log21021  3  log21024

Yes, 21.333 is an approximate solution.

Since 210  1024, x  1021 is a solution.

x  4

(b)

x  17

(b)

34  12  64

log217  3  log220

No, x  4 is not a solution.

Since 210  20, x  17 is not a solution.

x  64 3

(c)

364 3   64 Yes, x 

64 3

x  102  3  97

(c)

log297  3  log2100 Since 210  100, 102  3 is not a solution.

is a solution.

7. ln2x  3  5.8

8. lnx  1  3.8 x

(a)

1 2 3

 ln 5.8

x  1  e3.8

(a)

ln2 3  ln 5.8  3  lnln 5.8  5.8

ln1  e3.8  1  ln e3.8  3.8

No, x  12 3  ln 5.8 is not a solution.

Yes, x  1  e3.8 is a solution.

1 2

x  12 3  e5.8

(b)

x  45.701

(b)

ln2 3  e5.8  3  lne5.8  5.8

ln45.701  1  ln44.701  3.8

Yes, x  12 3  e5.8 is a solution.

Yes, x  45.701 is an approximate solution.

1 2

x  163.650

(c)

x  1  ln 3.8

(c)

ln2163.650  3  ln 330.3  5.8

ln1  ln 3.8  1  lnln 3.8  0.289

Yes, x  163.650 is an approximate solution.

No, x  1  ln 3.8 is not a solution.

9. 4x  16

10. 3x  243

11.

12 x  32

12.

14 x  64

4x  42

3x  35

2x  25

4x  43

x2

x5

x  5

x  3

x  5 13. ln x  ln 2  0 ln x  ln 2

ln x  ln 5

x2

x5

17. ln x  1 ln x

e

14. ln x  ln 5  0

e

1

18. ln x  7 ln x

e

e

7

15.

ex  2

x  3 16.

ex  4

ln ex  ln 2

ln e x  ln 4

x  ln 2

x  ln 4

x  0.693

x  1.386

19. log4 x  3 4log4 x



43

x  e1

x  e7

x  43

x  0.368

x  0.000912

x  64

20. log5 x  3 x  53 1

x  125 or 0.008

Section 3.4 21. f x  gx

22. f x  gx 27  9

2x  23

27x  2723 x

Point of intersection: 3, 8 2 2

25. e x  ex

Point of intersection:

2 8

e2 x  ex

26.

27.

2 2x

x  4x  2  0

2x 2  2x  0

x  log3 5 

x  0, x  1

2ex  10

x  log516 log 5 ln 5 or log 3 ln 3

x

4ex  91

33.

ex  9  19

ex  5

ex  91 4

ex  28

ln ex  ln 5

ln ex  ln 91 4

ln ex  ln 28

x  ln 5  1.609 34. 6x  10  47

35.

32x  80

36.

5x ln 6  ln 3000

2x ln 3  ln 80

ln 37 ln 6

5x 

ln 80 x  1.994 2 ln 3

x

x  2.015 37. 5t2  0.20 5t2 

1 5

5t2  51 t   1 2 t2 2x3  32 x  3  log2 32 x35 x8

65x  3000 ln 65x  ln 3000

ln 32x  ln 80

x  log6 37 x

x  ln 28  3.332

x  ln 91 4  3.125

6x  37

40.

ln 16 ln 5

x  1.723

x  1.465 32.

 ex2

5x  16

log3 3x  log3 5

2xx  1  0

2 3

30. 25x  32

43x  20 3x  5

x 2  x 2  2x

ex

By the Quadratic Formula x  1.618 or x  0.618.

x  2, x  4 29.

Point of intersection: 5, 0

x2  x  1  0

x 2  2x  8  0

x  1 or x  2 ex  ex

x5

x2  3  x  2

2x  x 2  8

0  x  1x  2

31.

x41

Point of intersection: 9, 2

23, 9

0  x2  x  2

2

elnx4 e0

x9

x  x2  2

28.

lnx  4  0

x  32

2 3

38.

43t  0.10 ln 43t  ln 0.10

39.

3x1  33 x13

ln 0.10 ln 4

x4

t

ln 0.10  0.554 3 ln 4

ln 3000 ln 6 ln 3000  0.894 5 ln 6

3x1  27

3t ln 4  ln 0.10 3t 

291

f x  gx

24.

log3 x  2

x

x3

f x  gx

23.

2 8 x

Exponential and Logarithmic Equations

292

Chapter 3

Exponential and Logarithmic Functions

23x  565

41.

82x  431

42.

ln 23x  ln 565

ln 82x  ln 431

3  x ln 2  ln 565

2  x ln 8  ln 431

3 ln 2  x ln 2  ln 565

2 ln 8  x ln 8  ln 431

x ln 2  ln 565  3 ln 2

x ln 8  ln 431  ln 82

x ln 2  3 ln 2  ln 565 x

x ln 8  ln 431  ln 64

3 ln 2  ln 565 ln 2

3

x

ln 565  6.142 ln 2

43. 8103x  12

44. 510x6  7

12 8

103x 

log 103x  log

10 x6 

32 

ln 5x1  ln 7 7 5

x  1 ln 5  ln 7 x1

ln 7 ln 5

x1

7 x  6  log 5



ln 7  2.209 ln 5

 6.146

 0.059 836x  40

47. e3x  12

36x  5

48.

x

6  x ln 3  ln 5

e2x  50 ln e2x  ln 50

3x  ln 12

ln 36x  ln 5

x 

35x1  21 5x1  7

7 x  6  log 5

3 1 x  log 3 2

6x

45.

7 5

log 10 x6  log

3 3x  log 2

46.

ln 431  ln 64  4.917 ln 8

2x  ln 50

ln 12  0.828 3

x

ln 5 ln 3

ln 50  1.956 2

ln 5 6 ln 3

x6

ln 5  4.535 ln 3

49. 500ex  300 ex  35 x  ln 35 x  ln 35  ln 53  0.511

50. 1000e4x  75 3 e4x  40 3 ln e4x  ln 40 3 4x  ln 40 3 x   14 ln 40

 0.648

51. 7  2ex  5

52. 14  3ex  11

2ex  2

3ex  25

ex  1

ex  25 3

x  ln 1  0

ln ex  ln 25 3 x  ln 25 3  2.120

Section 3.4 53. 623x1  7  9

log2 23x1

462x  3.5

8 3

6  2x  log4 3.5



3x  1 log2

55.

8462x  28

8  log2 3

x

6  2x 

83  loglog832  or lnln832 

1 log83  1  0.805 3 log 2

x3

ln 3.5  2.548 2 ln 4

ex  2ex  3  0 ex  5

(No solution)

57.

ln 3.5 ln 4

e2x  5ex  6  0

56.

ex  1ex  5  0 or

ln 3.5 ln 4

2x  6 

e2x  4ex  5  0 ex  1

ex  2 or ex  3

x  ln 5  1.609

x  ln 2  0.693 or x  ln 3  1.099

e2x  3ex  4  0

58. e2x  9ex  36  0

ex  1ex  4  0

ex2 9ex  36  0

ex  10 ⇒ ex  1

Because the discriminant is 92  4136  63, there is no solution.

Not possible since ex > 0 for all x. ex  40 ⇒ ex  4 ⇒ x  ln 4  1.386

59.

293

54. 8462x  13  41

623x1  16 23x1 

Exponential and Logarithmic Equations

500  20 100  e x2 500  20100  e x2

60.

400  350 1  ex 400  3501  ex

25  100  e x2

8  1  ex 7

e x2  75 x  ln 75 2

8  1  ex 7 1  ex 7

x  2 ln 75  8.635 ln

1  ln ex 7

x  ln

1 7

x  ln 71 x  ln 7 x  ln 7  1.946

61.

3000 2 2  e2x 3000  22  e2x 1500  2  e2x 1498  e2x ln 1498  2x x

ln 1498  3.656 2

294

62.

Chapter 3

Exponential and Logarithmic Functions

119 7 e  14

63.

6x

119  7e 6x  14



ln 1 

17  e 6x  14

0.065 365



31  e6x ln 31  ln

1  0.065 365  365t ln 1 

e 6x



365t

365t

64.



t

4  2.471 40 

9t

 21

65.



16  0.878 26  

3t ln 16 



12t

12t

 21.330

3t

3t

2  ln 2



0.10  ln 2 12

12t ln 1 

ln 21  0.247 9 ln 3.938225

t



0.10 12



9t ln 3.938225  ln 21

0.878 ln 16  26

1  0.10 12  ln 1 

ln 3.9382259t  ln 21



ln 4

365 ln1  0.065 365 

ln 31  0.572 6

3.9382259t  21

66.

 ln 4

0.065  ln 4 365

ln 31  6x x

4

t

 30

ln 2  6.960 12 ln1  0.10 12 

67. gx  6e1x  25 Algebraically:

 ln 30

15

6e1x  25



0.878  ln 30 26 t

6 −6

e1x 

ln 30  0.409 3 ln16  0.878 26 

25 6

1  x  ln

−30

256

x  1  ln

256

x  0.427 The zero is x  0.427. 68. f x  4ex1  15

69. f x  3e3x2  962

20

0  4ex1  15 15  4ex1 3.75 

ex1

ln 3.75  x  1 1  ln 3.75  x 1  ln 3.75  x 2.322  x The zero is 2.322.

Algebraically: −5

5

− 20

300 −6

9

3e3x2  962 e3x2 

962 3

−1200

 

3x 962  ln 2 3 x

 

2 962 ln 3 3

x  3.847 The zero is x  3.847.

Section 3.4 gx  8e2x3  11

70.

8e2x3  11

71. gt  e0.09t  3

5 −3

− 20

40

0.09t  ln 3

−15

x  0.478

−4

ln 3 0.09

t

x  1.5 ln 1.375

t  12.207

The zero is 0.478.

The zero is t  12.207.

72. f x  e1.8x  7

73. ht  e0.125t  8

e1.8x  7  0

Algebraically:

74. f x  e2.724x  29 e2.724x  29

e0.125t  80

e1.8x  7

2.724x  ln 29

e0.125t  8

e1.8x  7

x

0.125t  ln 8

1.8x  ln 7 x

8

e0.09t  3

2x  ln 1.375 3

295

Algebraically:

7

e2x3  1.375 

Exponential and Logarithmic Equations

ln 7 1.8

t

x  1.236

ln 8 0.125

The zero is 1.236.

t  16.636

x  1.081

10

The zero is t  16.636.

The zero is 1.081. 13

ln 29 2.724

−5

5

2 −40

40

−35 −5

5

−7

−10

75. ln x  3

76. ln x  2

x  e3  0.050

77. ln 2x  2.4 2x 

eln x  e2 x  e2  7.389

78. ln 4x  1

e2.4

eln 4x  e1

e2.4  5.512 x 2

4x  e x

80. log 3z  2

79. log x  6

81. 3 ln 5x  10

10log 3z  102

x  106  1,000,000.000

3z  100 100 z  33.333 3

83. ln x  2  1 x  2  e1

x2

e2

x  e2  2  5.389

84. ln x  8  5 eln x8  e5 x  8 

e5

x  8  e10 x  e10  8  22,034.466

ln 5x 

82. 2 ln x  7

10 3

ln x 

5x  e103 e103 x  5.606 5 85. 7  3 ln x  5 3 ln x  2 ln x 

e  0.680 4

 23

7 2

eln x  e72 x  e72  33.115 86. 2  6 ln x  10 6 ln x  8 ln x   43

x  e23

eln x  e43

 0.513

x  e43  0.264

296

Chapter 3

Exponential and Logarithmic Functions

87. 6 log30.5x  11

88. 5 log10x  2  11

log30.5x  11 6

log10x  2  11 5

3log30.5x  3116

10log10x2  10115

0.5x  3116

x  2  10115

x  23116  14.988

x  10115  2  160.489

89. ln x  lnx  1  2

90. ln x  lnx  1  1

x  1  2

lnxx  1  1

x  e2 x1

xx  1  e1

ln

x

elnxx1  e1 x2  x  e  0

x  e2x  1

x

x  e2x  e2

1 ± 1  4e 2

x  e2x  e2 The only solution is x 

x1  e2  e2 x

1  1  4e  1.223. 2

e2  1.157 1  e2

This negative value is extraneous. The equation has no solution. 91. ln x  lnx  2  1

92. ln x  lnx  3  1

lnxx  2  1

lnxx  3  1

xx  2 

e1

elnxx3  e1

x2  2x  e  0

xx  3  e1

x

2 ± 4  4e 2

x2  3x  e  0 x

2 ± 2 1  e   1 ± 1  e 2

The only solution is x 

The negative value is extraneous. The only solution is x  1  1  e  2.928. ln x  5  lnx  1  lnx  1

93.

lnx  5  ln x5

x1

x  1

x1 x1

x  5x  1  x  1 x2

 6x  5  x  1

x2  5x  6  0

x  2x  3  0 x  2 or x  3 Both of these solutions are extraneous, so the equation has no solution.

3 ± 9  4e 2

94.

3  9  4e  0.729. 2

lnx  1  lnx  2  ln x ln

x1

x  2  ln x x1 x x2 x  1  x2  2x 0  x2  3x  1

 3 ± 32  411 x 21 3 ± 13 x 2 3.303  x (The negative apparent solution is extraneous.)

95. log22x  3  log2x  4 2x  3  x  4 x7

Section 3.4

Exponential and Logarithmic Equations

297

96. logx  6  log2x  1 x  6  2x  1 7  x The apparent solution x  7 is extraneous, because the domain of the logarithm function is positive numbers, and 7  6 and 27  1 are negative. There is no solution. 97. logx  4  log x  logx  2

98. log2 x  log2x  2  log2x  6

x4  logx  2 x

log2xx  2  log2x  6

log





xx  2  x  6 x2

x4 x2 x x  4  x2  2x 0

x2

x60

x  3x  2  0 x  3 or x  2

x4

1 ± 17 x 2

The value x  3 is extraneous. The only solution is x  2.

Choosing the positive value of x (the negative value is extraneous), we have x

1  17  1.562. 2 1 2 1 x log4  x1 2

100. log3 x  log3x  8  2

99. log4 x  log4x  1 



log3xx  8  2



3log3x

2 8x

 32

x2  8x  9

4log4xx1  412 x  412 x1

x2  8x  9  0

x  9x  1  0

x  2x  1

x  9 or x  1

x  2x  2

The value x  1 is extraneous. The only solution is x  9.

x  2 x2 101. log 8x  log1  x  2 log

8x 2 1  x 8x  102 1  x

8x  1001  x 

2x  251  x   25  25 x 2x  25  25 x

2x  252  25 x

2

4x2  100x  625  625x 4x2  725x  625  0 x

725 ± 7252  44625 2529 ± 5 33 725 ± 515,625   24 8 8

x  0.866 (extraneous) or x  180.384 The only solution is x 

2529  5 33  180.384. 8

298

Chapter 3

Exponential and Logarithmic Functions

102. log 4x  log12  x   2 log

12 4x x  2

10log4x (12 x   102 4x  100 12  x 4x  10012  x  4x  1200  100 x 4x  1200  100 x x  300  25 x

x  3002  25 x 

2

x2  600x  90,000  625x x2  1225x  90,000  0 x

1225 ± 12252  4190,000 2

x

1225 ± 1,140,625 2

x

1225 ± 125 73 2

x  78.500 extraneous or x  1146.500 The only solution is x 

1225  125 73  1146.500. 2

103. y1  7 y2 

104.

10

2x

From the graph we have x  2.807 when y  7. Algebraically:

ln

−8

10

ln

−2

2 ln

 ln 7

105. y1  3

x  e 3  20.086

18

4 lnx  2  10

y2  ln x

ln x  3

1 x 3

106. 10  4 lnx  2  0

5

3  ln x  0

10 − 200

The solution is x  2.197.

ln 7  2.807 ln 2

From the graph we have x  20.086 when y  3. Algebraically:

−2

1 x  3 2

2.197  x

x ln 2  ln 7 x

800

1  ex2 3

2x  7 2x

500  1500ex2

−5

30 −1

lnx  2  2.5 elnx2  e2.5 x  2  e2.5 x  e2.5  2 x  14.182 The solution is x  14.182.

−5

30 −3

Section 3.4 A  Pert

107. (a)

5000 

A  Pert

(b)

Exponential and Logarithmic Equations r  0.12

108. (a)

3  e0.085t

ln 2  0.085t

5000  2500e0.12t

7500  2500e0.12t

2  e0.12t

3  e0.12t

ln 2  ln e0.12t

ln 3  ln e0.12t

ln 2  0.12t

ln 3  0.12t

ln 3  0.085t

ln 2 t 0.085

ln 3 t 0.085

t  8.2 years

A  Pert

rt

0.085t

2  e0.085t

r  0.12

(b)

A  Pe

7500  2500e

2500e0.085t

t  12.9 years

299

ln 2 t 0.12

ln 3 t 0.12

t  5.8 years

t  9.2 years

109. p  500  0.5e0.004x p  350

(a)

(b)

p  300

350  500  0.5e0.004x

300  500  0.5e0.004x

300  e0.004x

400  e0.004x

0.004x  ln 300

0.004x  ln 400

x  1426 units



110. p  5000 1 

x  1498 units

4 4  e0.002x



(a) When p  \$600:

(b) When p  \$400:



600  5000 1  0.12  1 

4 4  e0.002x





400  5000 1 

4 4  e0.002x

0.08  1 

4  0.88 4  e0.002x



4 4  e0.002x

4  0.92 4  e0.002x

4  3.52  0.88e0.002x

4  3.68  0.92e0.002x

0.48  0.88e0.002x

0.32  0.92e0.002x

6  e0.002x 11

8  e0.002x 23

ln

6  ln e0.002x 11

ln

8  ln e0.002x 23

ln

6  0.002x 11

ln

8  0.002x 23

x

4 4  e0.002x

ln611  303 units 0.002

x

ln823  528 units 0.002

111. V  6.7e48.1t , t ≥ 0 (a)

(b) As t → , V → 6.7.

10

1.3  6.7e48.1t

(c)

Horizontal asymptote: V  6.7

0

1500 0

The yield will approach 6.7 million cubic feet per acre.

1.3  e48.1t 6.7 ln

67  13

t

48.1 t 48.1  29.3 years ln1367

300

Chapter 3

Exponential and Logarithmic Functions

112. N  68100.04x

113. y  7312  630.0 ln t, 5 ≤ t ≤ 12

When N  21:

7312  630.0 ln t  5800

21  6810

0.04x



630.0 ln t  1512

21  100.04x 68

ln t  2.4 t  e2.4  11

21 log10  0.04x 68 x

t  11 corresponds to the year 2001.

log102168  12.76 inches 0.04

114. y  4381  1883.6 ln t, 5 ≤ t ≤ 13 9000  4381  1883.6 ln t 4619  1883.6 ln t ln t 

4619  2.45222 1883.6

t  e2.45222  11.6 Since t  5 represents 1995, t  11.6 indicates that the number of daily fee golf facilities in the U.S. reached 9000 in 2001. 115. (a) From the graph shown in the textbook, we see horizontal asymptotes at y  0 and y  100. These represent the lower and upper percent bounds; the range falls between 0% and 100%. Females

(b) Males 50 

100 1  e0.6114x69.71

50 

1  e0.6114x69.71  2

1  e0.66607x64.51  2

e0.6114x69.71  1

e0.6667x64.51  1

0.6114x  69.71  ln 1

0.66607x  64.51  ln 1

0.6114x  69.71  0

0.66607x  64.51  0 x  64.51 inches

x  69.71 inches

116. P  (a)

100 1  e0.66607x64.51

0.83 1  e0.2n (c) When P  60% or P  0.60:

1.0

0.60  0

40 0

(b) Horizontal asymptotes: P  0, P  0.83 The upper asymptote, P  0.83, indicates that the proportion of correct responses will approach 0.83 as the number of trials increases.

1  e0.2n  e0.2n 

0.83 1  e0.2n 0.83 0.60 0.83 1 0.60

ln e0.2n  ln

0.60  1

0.2n  ln

0.60  1

0.83 0.83

ln n

0.60  1 0.83

0.2

 5 trials

Section 3.4

117. y  3.00  11.88 ln x  (a)

36.94 x

Exponential and Logarithmic Equations

118. T  201  72h (a) From the graph in the textbook we see a horizontal asymptote at T  20. This represents the room temperature.

x

0.2

0.4

0.6

0.8

1.0

y

162.6

78.5

52.5

40.5

33.9

100  201  72h

(b) (b)

301

5  1  72h

200

4  72h 0

4  2h 7

1.2 0

The model seems to fit the data well.

ln

7  ln 2

ln

7  h ln 2

(c) When y  30: 36.94 30  3.00  11.88 ln x  x

4

h

4

ln47 h ln 2

Add the graph of y  30 to the graph in part (a) and estimate the point of intersection of the two graphs. We find that x  1.20 meters.

h  0.81 hour

(d) No, it is probably not practical to lower the number of gs experienced during impact to less than 23 because the required distance traveled at y  23 is x  2.27 meters. It is probably not practical to design a car allowing a passenger to move forward 2.27 meters (or 7.45 feet) during an impact. 120. logau  v  loga uloga v

119. logauv  loga u  loga v

False.

True by Property 1 in Section 3.3.

2.04  log1010  100  log10 10log10 100  2

121. logau  v  loga u  loga v

122. loga

False.

uv  log

a

u  loga v

123. Yes, a logarithmic equation can have more than one extraneous solution. See Exercise 93.

True by Property 2 in Section 3.3.

1.95  log100  10  log 100  log 10  1 124. A  Pert

125. Yes.

(a) A  2P

ert

 2





Pert



2P  Pe

4P  Pert

(c) A  Per2t  Pertert  ertPert

2  ert

4  ert

ln 2  rt

ln 4  rt

(b) A 

Pertert



Time to Double

Pert

Pe2rt

ert

Doubling the interest rate yields the same result as doubling the number of years. If 2 > ert (i.e., rt < ln 2), then doubling your investment would yield the most money. If rt > ln 2, then doubling either the interest rate or the number of years would yield more money.

rt

ln 2 t r

2 ln 2 t r

Thus, the time to quadruple is twice as long as the time to double.

302

Chapter 3

Exponential and Logarithmic Functions

126. (a) When solving an exponential equation, rewrite the original equation in a form that allows you to use the One-to-One Property ax  ay if and only if x  y or rewrite the original equation in logarithmic form and use the Inverse Property loga ax  x.

128. 32  2 25  16

127. 48x2y5  16x2y43y

3



10  2

10  2

3 10  2

 2  25

3 3 3 25 15  375 129. 3 3  125  3  5 3

 4 2  10

 4 x y 2 3y

130.

(b) When solving a logarithmic equation, rewrite the original equation in a form that allows you to use the One-to-One Property loga x  loga y if and only if x  y or rewrite the original equation in exponential form and use the Inverse Property aloga x  x.

 10  2

131. f x  x  9

y

3 10  2  10  4

Domain: all real numbers x

3 10  2  6

y-axis symmetry



8 6 4

y

2

12

y-intercept: 0, 9

x

10  2

14

±1

0 9

10

±2

11

2

±3

x

−8 −6 − 4 − 2 −2

12

2

4

6

8

1

3

4

1  10  1 2

133. gx 

y

132. 8 6

2x, x  4, 2

x < 0 x ≥ 0

y 5

Domain: all real numbers x

4

4

3

x-intercept: 2, 0

2 x

−6 − 4 − 2 −2

2

4

6

8

2 1

y-intercept: 0, 4

x

−4 −3 − 2 − 1

−4 −6

−3

x

3

2

1

0.5

0

1

2

3

y

6

4

2

1

4

3

2

5

y

134. 6 4 1 −6

−4

−2

x 2

4

6

−2

−6

135. log6 9 

log10 9 ln 9   1.226 log10 6 ln 6

137. log34 5 

log10 5 ln 5   5.595 log1034 ln34

136. log3 4 

log10 4 ln 4   1.262 log10 3 ln 3

138. log8 22 

log10 22 ln 22   1.486 log10 8 ln 8

Section 3.5

Section 3.5 ■

Exponential and Logarithmic Models

303

Exponential and Logarithmic Models

You should be able to solve growth and decay problems. (a) Exponential growth if b > 0 and y  aebx. (b) Exponential decay if b > 0 and y  aebx.

You should be able to use the Gaussian model y  aexb c. 2

You should be able to use the logistic growth model a . y 1  berx

You should be able to use the logarithmic models y  a  b ln x, y  a  b log x.

Vocabulary Check 1. y  aebx; y  aebx

2. y  a  b ln x; y  a  b log x

4. bell; average value

5. sigmoidal

1. y  2ex4

3. normally distributed

3. y  6  logx  2

2. y  6ex4

This is an exponential growth model. Matches graph (c).

4. y  3ex2 5 2

This is a Gaussian model. Matches graph (a).

This is a logarithmic function shifted up six units and left two units. Matches graph (b).

This is an exponential decay model. Matches graph (e).

6. y 

5. y  lnx  1 This is a logarithmic model shifted left one unit. Matches graph (d).

7. Since A  1000e0.035t, the time to double is given by 2000  1000e0.035t and we have

1500  750e0.105t

ln 2  ln e0.035t

2  e0.105t

t

ln 2  19.8 years. 0.035

Amount after 10 years: A  1000e0.35  \$1419.07

This is a logistic growth model. Matches graph (f).

8. Since A  750e0.105t, the time to double is given by 1500  750e0.105t, and we have

2  e0.035t ln 2  0.035t

4 1  e2x

ln 2  ln e0.105t ln 2  0.105t t

ln 2  6.60 years. 0.105

Amount after 10 years: A  750e0.10510  \$2143.24

304

Chapter 3

Exponential and Logarithmic Functions

9. Since A  750ert and A  1500 when t  7.75, we have the following.

10. Since A  10,000ert and A  20,000 when t  12, we have

1500  750e7.75r

20,000  10,000e12r

2  e7.75r

2  e12r

ln 2  ln e7.75r

ln 2  ln e12r

ln 2  7.75r

ln 2  12r

r

ln 2  0.089438  8.9438% 7.75

r

Amount after 10 years: A  750e0.08943810  \$1834.37

ln 2  0.057762  5.7762%. 12

Amount after 10 years: A  10,000e0.05776210  \$17,817.97

11. Since A  500ert and A  \$1505.00 when t  10, we have the following. 1505.00  r

12. Since A  600ert and A  19,205 when t  10, we have 19,205  600e10r

500e10r

19,205  e10r 600

ln1505.00500  0.110  11.0% 10

The time to double is given by 1000  500e0.110t t

ln 2  6.3 years. 0.110

ln

 ln e 19,205 600 

ln

 10r 19,205 600 

10r

r

ln19,205600  0.3466 or 34.66%. 10

The time to double is given by 1200  600e0.3466t t

13. Since A  Pe0.045t and A  10,000.00 when t  10, we have the following. 10,000.00 

14. Since A  Pe0.02t and A  2000 when t  10, we have 2000  Pe0.0210

Pe0.04510

P

10,000.00  P  \$6376.28 e0.04510 The time to double is given by t 



15. 500,000  P 1  P



0.075 12



500,000 0.075 1220 1 12



500,000   \$112,087.09 1.00625240

2000  \$1637.46. e0.0210

The time to double is given by t 

ln 2  15.40 years. 0.045

1220

ln 2  2 years. 0.3466

16.



AP 1



500,000  P 1 

r n



nt

0.12 12

P  \$4214.16



12(40)

ln 2  34.7 years. 0.02

Section 3.5

Exponential and Logarithmic Models

305

17. P  1000, r  11% n1

(a)

n  12

(b)

1  0.11t  2

1  0.11 12 

t ln 1.11  ln 2 t



12t ln 1 

ln 2  6.642 years ln 1.11

1  0.11 365  

365t ln 1 

365t



t

2

ln 2

12 ln1  0.11 12 

(d) Compounded continuously e0.11t  2



0.11  ln 2 365

0.11t  ln 2 ln 2

t

2

0.11  ln 2 12

n  365

(c)

12t

365 ln1 

0.11 365



t

 6.302 years

ln 2  6.301 years 0.11

18. P  1000, r  10.5%  0.105 (b) n  12

(a) n  1 ln 2  6.94 years ln1  0.105

t

t

(c) n  365 365 ln1  0.105 365 

3P  Pert

19.

12 ln1  0.105 12 

 6.63 years

(d) Compounded continuously ln 2

t

ln 2

t

 6.602 years

r

3  ert t

ln 3  rt

ln 3 (years) r

ln 2  6.601 years 0.105

2%

4%

6%

8%

10%

12%

54.93

27.47

18.31

13.73

10.99

9.16

ln 3 t r 20.

60

0

0.16 0

Using the power regression feature of a graphing utility, t  1.099r1. 21.

3P  P1  rt

r

3  1  rt ln 3  ln1  rt ln 3  t ln1  r ln 3 t ln1  r

t

ln 3 (years) ln1  r

2%

4%

6%

8%

10%

12%

55.48

28.01

18.85

14.27

11.53

9.69

 6.330 years

306

Chapter 3

22.

Exponential and Logarithmic Functions 23. Continuous compounding results in faster growth.

60

A  1  0.075 t  and A  e0.07t A 0.16

Amount (in dollars)

0 0

Using the power regression feature of a graphing utility, t  1.222r1.

A = e0.07t

2.00 1.75 1.50 1.25

A = 1 + 0.075 [[ t [[

1.00

t

2

4

6

8

10

Time (in years)

24. 2

(

1 C  Cek1599 2

25.

)

0.055 [[365t [[

A = 1 + 365

26.

1 C  Cek1599 2

0.5  ek1599 ln 0.5  ln 0

ln 0.5  k1599

10 0

1  ek1599 2

ek1599

A = 1 + 0.06 [[ t [[

k

512%

From the graph, compounded daily grows faster than 6% simple interest.

ln 0.5 1599

Given C  10 grams after 1000 years, we have

ln

1  ln ek1599 2

ln

1  k1599 2 k

y  10e ln 0.51599 1000

ln12 1599

Given y  1.5 grams after 1000 years, we have

 6.48 grams.

1.5  Ce ln121599 1000 C  2.31 grams.

27.

1 C  Cek5715 2

28.

1 C  Cek5715 2

0.5  ek5715

1  ek5715 2

ln 0.5  ln ek5715 ln 0.5  k5715 k

ln 0.5 5715

Given y  2 grams after 1000 years, we have 2  Ce ln 0.55715 1000 C  2.26 grams.

ln

1  ln ek5715 2

ln

1  k5715 2 k

ln12 5715

Given C  3 grams, after 1000 years we have y  3e ln125715 1000 y  2.66 grams.

29.

1 C  Cek24,100 2 0.5  ek24,100 ln 0.5  ln ek24,100 ln 0.5  k24,100 k

ln 0.5 24,100

Given y  2.1 grams after 1000 years, we have 2.1  Ce ln 0.524,100 1000 C  2.16 grams.

Section 3.5

30.

1 C  Cek24,100 2

y  aebx

31.

1  ln ek24,100 2

ln

1  k24,100 2

1 1  aeb0 ⇒ a  2 2

10  eb3

1 5  eb4 2

ln 10  3b ln 10  b ⇒ b  0.7675 3

10  e4b ln 10  ln e4b

Thus, y  e0.7675x .

ln12 k 24,100

307

y  aebx

32.

1  aeb0 ⇒ 1  a

1  ek24,100 2 ln

Exponential and Logarithmic Models

ln 10  4b

Given y  0.4 grams after 1000 years, we have

ln 10  b ⇒ b  0.5756 4

0.4  Ce ln1224,100 1000

Thus, y  12e0.5756x.

C  0.41 grams. y  aebx

33.

ln

y  aebx

34.

5  aeb0 ⇒ 5  a

1  aeb0 ⇒ 1  a

1  5eb4

1  eb3 4

1  e4b 5

ln

15  4b

14  ln e

ln

4  3b

ln15  b ⇒ b  0.4024 4

3b

1

ln14 b 3

Thus, y  5e0.4024x.

⇒ b  0.4621

Thus, y  e0.4621x .

35. P  2430e0.0029t (a) Since the exponent is negative, this is an exponential decay model. The population is decreasing.

(c) 2.3 million  2300 thousand 2300  2430e0.0029t

(b) For 2000, let t  0: P  2430 thousand people

2300  e0.0029t 2430

For 2003, let t  3: P  2408.95 thousand people ln

 0.0029t 2300 2430  t

ln23002430  18.96 0.0029

The population will reach 2.3 million (according to the model) during the later part of the year 2018. 36.

Country

2000

2010

Bulgaria

7.8

7.1

31.3

34.3

1268.9

1347.6

59.5

61.2

282.3

309.2

China United Kingdom United States —CONTINUED—

308

Chapter 3

Exponential and Logarithmic Functions

(a) Bulgaria:

a  31.3

a  7.8

34.3  31.3eb10

7.1  7.8eb10 ln

7.1  10b ⇒ b  0.0094 7.8

ln

34.3  10b ⇒ b  0.00915 31.3

For 2030, use t  30.

For 2030, use t  30.

y  7.8e0.009430  5.88 million

y  31.3e0.0091530  41.2 million United States:

China:

ln

a  1268.9

a  282.3

1347.6  1268.9eb10

309.2  282.3eb10

1347.6  10b ⇒ b  0.00602 1268.9

ln

309.2  10b ⇒ b  0.0091 282.3

For 2030, use t  30.

For 2030, use t  30.

y  1268.9e0.0060230  1520.06 million

y  282.3e0.009130  370.9 million

United Kingdom: a  59.5 61.2  59.5eb10 ln

61.2  10b ⇒ b  0.00282 59.5

For 2030, use t  30. y  59.5e0.0028230  64.7 million (b) The constant b determines the growth rates. The greater the rate of growth, the greater the value of b. (c) The constant b determines whether the population is increasing b > 0 or decreasing b < 0. 37. y  4080ekt

y  10ekt

38.

65  10ek14

When t  3, y  10,000: 10,000  4080ek3 10,000  e3k 4080 ln

 3k 10,000 4080  k

ln10,0004080  0.2988 3

When t  24: y  4080e0.298824  5,309,734 hits

ln

 14k ⇒ k  0.1337 65 10 

For 2010, t  20: y  10e0.133720  \$144.98 million

Section 3.5 39.

N  100ekt

Exponential and Logarithmic Models

N  250ekt

40.

280  250ek10

300  100e5k 3  e5k

1.12  e10k

ln 3  ln e5k

k

ln 3  5k

500  250e ln 1.1210 t 2  e ln 1.1210 t

N  100e0.2197t

ln 2 

200  100e0.2197t

41. R 

ln 2  3.15 hours 0.2197

t

1 t8223 e 1012 R

(a)



ln

1  5715k 2

1012 814

k

 

t 1012  ln 14 8223 8

ln12 5715

The ancient charcoal has only 15% as much radioactive carbon. 0.15C  Ce ln 0.55715 t

108   12,180 years old 12

14

ln 0.15 

1 t8223 1 e  11 (b) 1012 13

t

1012 et8223  11 13 

ln 2  61.16 hours ln 1.1210

1 C  Ce5715k 2

1 814

t  8223 ln

ln 101.12t

y  Cekt

42.

1 t8223 1 e  14 1012 8 et8223 

ln 1.12 10

N  250e ln 1.1210 t

ln 3 k  0.2197 5

t

ln 0.5 t 5715 5715 ln 0.15  15,642 years ln 0.5

 

t 1012  ln 8223 1311

t  8223 ln

 4797 years old 10 13  12 11

43. 0, 30,788, 2, 18,000 (a) m 

309

18,000  30,788  6394 20

a  30,788

(b)

32,000

18,000  30,788ek2

b  30,788

4500  e2k 7697

Linear model: V  6394t  30,788 ln

0

4 0

 2k 4500 7697  k

—CONTINUED—

(c)





4500 1 ln  0.268 2 7697

Exponential model: V  30,788e0.268t

The exponential model depreciates faster in the first two years.

310

Chapter 3

Exponential and Logarithmic Functions

43. —CONTINUED— (d)

t

1

3

V  6394t  30,788

\$24,394

\$11,606

V  30,788e

\$23,550

\$13,779

0.268t

(e) The linear model gives a higher value for the car for the first two years, then the exponential model yields a higher value. If the car is less than two years old, the seller would most likely want to use the linear model and the buyer the exponential model. If it is more than two years old, the opposite is true.

44. 0, 1150, 2, 550 (a) m 

550  1150  300 20

V  300t  1150 (c)

550  1150ek2

(b) ln

550 1150   2k ⇒ k  0.369

1200

V  1150e0.369t (d)

0

4 0

The exponential model depreciates faster in the first two years.

t

1

3

V  300t  1100

\$850

\$250

V  1150e0.369t

\$795

\$380

(e) The slope of the linear model means that the computer depreciates \$300 per year, then loses all value in the third year. The exponential model depreciates faster in the first two years but maintains value longer. 45. St  1001  ekt 15  1001  ek1

(b)

85  100ek 85 100

 ek

0.85  ek ln 0.85  ln

S

Sales (in thousands of units)

(a)

ek

120 90 60 30 t 5 10 15 20 25 30

Time (in years)

k  ln 0.85 k  0.1625

(c) S5  1001  e0.16255  55.625  55,625 units

St  1001  e0.1625t 46. N  301  ekt (a)

N  19, t  20

N  25

(b)

25  301  e0.050t

19  301  e20k 30e20k  11 e20k 

11 30

 

11 ln e20k  ln 30 20k  ln

11 30 

k  0.050 So, N  301  e0.050.

5  e0.050t 30 ln

305   ln e

ln

305   0.050t

0.050t

t

ln530  36 days 0.050

Section 3.5 47. y  0.0266ex100 450, 70 ≤ x ≤ 116 2

(a)

Exponential and Logarithmic Models

48. (a)

311

0.9

0.04

4

7 0

70

115 0

(b) The average IQ score of an adult student is 100.

49. pt 

1000 1  9e0.1656t

(a) p5 

50. S 

1000  203 animals 1  9e0.16565 500 

(b)

(b) The average number of hours per week a student uses the tutor center is 5.4.

1000 1  9e0.1656t

(a)

500,000 1  0.6ekt 300,000 

1  0.6e4k 

5 3

0.6e4k 

2 3

1  9e0.1656t  2 9e0.1656t  1 e0.1656t

e4k 

1  9

k

1200

So, S  0

40 0

The horizontal asymptotes are p  0 and p  1000. The asymptote with the larger p-value, p  1000, indicates that the population size will approach 1000 as time increases.

51. R  log

I  log I since I0  1. I0

10 9

4k  ln

ln19 t  13 months 0.1656 (c)

500,000 1  0.6e4k

9 10

 

1 10 ln  0.0263 4 9

500,000 . 1  0.6e0.0263t

(b) When t  8: S

52. R  log

500,000  287,273 units sold. 1  0.6e0.02638

I  log I since I0  1. I0

(a) 7.9  log I ⇒ I  107.9  79,432,823

(a) R  log 80,500,000  7.91

(b) 8.3  log I ⇒ I  108.3  199,526,231

(b) R  log 48,275,000  7.68

(c) 4.2  log I ⇒ I  104.2  15,849

(c) R  log 251,200  5.40

53.   10 log

I where I0  1012 wattm2. I0

(a)   10 log

1010  10 log 102  20 decibels 1012

(b)   10 log

105  10 log 107  70 decibels 1012

(c)   10 log

108  10 log 104  40 decibels 1012

(d)   10 log

1  10 log 1012  120 decibels 1012

312

Chapter 3

54. I  10 log

Exponential and Logarithmic Functions

I where I0  1012 wattm2 I0

(a) 1011  10 log

  10 log

(b) 102  10 log

104  10 log 108  80 decibels 1012

(c) 104  10 log

55.

1011  10 log 101  10 decibels 1012

I I0

(d) 102  10 log

56.

 I  log 10 I0

102  10 log 1010  100 decibels 1012

  10 log10 1010 

10 10  10log II0 1010 

102  10 log 1014  140 decibels 1012

I I0

I I0

I  I01010

I I0

% decrease 

I0108.8  I0107.2  100  97% I0108.8

I  I010 10 % decrease 

I0109.3  I0108.0  100  95% I0109.3

57. pH  log H

58. pH  log H

log2.3  105  4.64 59.

5.8  log H

log 11.3  106  4.95 60.

5.8  log H

3.2  log H

103.2  H



105.8  10log H

H  6.3  104 mole per liter

105.8  H

H  1.58  106 mole per liter 61.

2.9  log H

2.9  log H

H  102.9 for the apple juice 8.0  log H

8.0  log H

H  108 for the drinking water 102.9 108

 105.1 times the hydrogen ion concentration of drinking water

63. t  10 ln

T  70 98.6  70

At 9:00 A.M. we have: t  10 ln

85.7  70  6 hours 98.6  70

From this you can conclude that the person died at 3:00 A.M.

62.

pH  1  log H

  pH  1  log H

10pH1  H

10pH1  H

10pH  10  H

The hydrogen ion concentration is increased by a factor of 10.

Section 3.5



Pr 12

64. Interest: u  M  M 



Principal: v  M 

Pr 12

1  12 r

1  12 r

313

12t

12t

(a) P  120,000, t  35, r  0.075, M  809.39

(c) P  120,000, t  20, r  0.075, M  966.71

800

800

u

u

v

v 0

35

0

0

20 0

(b) In the early years of the mortgage, the majority of the monthly payment goes toward interest. The principal and interest are nearly equal when t  26 years.

65. u  120,000

(a)

Exponential and Logarithmic Models

0.075t 1 1 1  0.07512





12t

1

150,000

0

The interest is still the majority of the monthly payment in the early years. Now the principal and interest are nearly equal when t  10.729  11 years.

24

(b) From the graph, u  \$120,000 when t  21 years. It would take approximately 37.6 years to pay \$240,000 in interest. Yes, it is possible to pay twice as much in interest charges as the size of the mortgage. It is especially likely when the interest rates are higher.

0

66. t1  40.757  0.556s  15.817 ln s t2  1.2259  0.0023s2 (a) Linear model: t3  0.2729s  6.0143 Exponential model: t4  1.5385e0.02913s or t4  1.53851.0296s (b)

t2

25

t4 t3

20

t1

100 0

(c)

s

30

40

50

60

70

80

90

t1

3.6

4.6

6.7

9.4

12.5

15.9

19.6

t2

3.3

4.9

7.0

9.5

12.5

15.9

19.9

t3

2.2

4.9

7.6

10.4

13.1

15.8

18.5

t4

3.7

4.9

6.6

8.8

11.8

15.8

21.2

Note: Table values will vary slightly depending on the model used for t4.

S2  3.4  3.3  5  4.9  7  7  9.3  9.5  12  12.5 

15.8  15.9  20  19.9  1.1 S3  3.4  2.2  5  4.9  7  7.6  9.3  10.4  12  13.1 

15.8  15.8  20  18.5  5.6 S4  3.4  3.7  5  4.9  7  6.6  9.3  8.9  12  11.9 

15.8  15.9  20  21.2  2.6

(d) Model t1: S1  3.4  3.6  5  4.6  7  6.7  9.3  9.4  12  12.5  15.8  15.9  20  19.6  2.0 Model t2: Model t3: Model t4:

The quadratic model, t2, best fits the data.

314

Chapter 3

Exponential and Logarithmic Functions

67. False. The domain can be the set of real numbers for a logistic growth function.

68. False. A logistic growth function never has an x-intercept.

69. False. The graph of f x is the graph of gx shifted upward five units.

70. True. Powers of e are always positive, so if a > 0, a Gaussian model will always be greater than 0, and if a < 0, a Gaussian model will always be less than 0.

71. (a) Logarithmic

(b) Logistic (c) Exponential (decay) (d) Linear (e) None of the above (appears to be a combination of a linear and a quadratic) (f) Exponential (growth) 73. 1, 2, 0, 5

74. 4, 3, 6, 1 y

(a)

y

(a) (0, 5)

5

6 4

(− 6, 1)

3 2

(− 1, 2)

−6

2 x

−4

2

−2

−1

2

3

−1

−6

(b) d  0  12  5  22  12  32  10 (c) Midpoint:

(b) d  6  42  1  32

12 0, 2 2 5   21, 72

 100  16  116  229 (c) Midpoint:

3 52  3 (d) m  0  1 1

76. 10, 4, 7, 0

y

y

(a)

8

6

(10, 4)

6 4

4

(3, 3)

2

2 −2 −2

x 2

4

6

8 10

14

(14, − 2)

−4

(7, 0) −2

2

4

6

x 8

10

−2 −4

−6

−6

−8

(b) d  14  32  2  32  112  52  146 (c) Midpoint: (d) m 

62 4, 32 1  1, 1

4 3  1 2   4  6 10 5

(d) m 

75. 3, 3, 14, 2 (a)

(4, −3)

−4

x 1

6

−2

1 −3

4

3 2 14, 3  22  172, 12

5 2  3  14  3 11

(b) d  10  72  4  02  9  16  25  5 (c) Midpoint: (d) m 

7 2 10, 0 2 4  172, 2

4 40  10  7 3

Section 3.5

77.

12,  41, 34, 0

78.

y

(a)

Exponential and Logarithmic Models

315

73, 16,  32,  31 y

(a) 2

1

1

1 2

( ( 3 ,0 4

(

1 , −1 2 4

(

3

3

( 2

2

3

(

 32  37   31  61 1  3     9.25 2

2

2

2

2

14 0  14  1 (d) m  34  12 14

79. y  10  3x



232 73, 132 16  56,  121  13  16 12 1   23  73 3 6

80. y  4x  1

y 3

Line

10

2

Slope: m  4

8 6

y-intercept: 0, 10

y-intercept: 0, 1

4

−3

−2

1

2

3

−1 −2

x

−2 −2

2

6

81. y  2x2  3

8

10 12

−3

82. y  2x2  7x  30

y

y  2x  02  3

2

−6

x

−1

2

Parabola

2

(c) Midpoint:

(d) m 

y

Slope: m  3

2

(b) d 

12 2 34, 142  0  58,  81

Line

2

−2

34  21  0   41 1 1 1         4 4 8

(b) d 

(c) Midpoint:

1

− 2, − 1

2

−1 2

x

−1

x

1

−1

( 73 , 16 (

−4

−2

x 2

4

6

−2

Vertex: 0, 3

 2x  5x  6  2x  4   7 2

y x

−4

2

−5

289 8

4

8

Parabola

74,  2898  5 x-intercepts:  2, 0, 6, 0 Vertex:

83. 3x2  4y  0 3x2  4y 4 3y

5

x2 

Parabola

4 3 2

Vertex: 0, 0

1

1 Focus: 0, 3 

− 4 −3 − 2 − 1

1

Directrix: y   3

y

x2  8y

6

Parabola

− 35

84. x2  8y  0

y 7

− 30

x 1

2

3

4

2

−6

−4

x

4 −2

Vertex: 0, 0

−4

Focus: 0, 2

−6

Directrix: y  2

−8 − 10

6

316

Chapter 3

85. y 

Exponential and Logarithmic Functions

4 1  3x

86. y 

Vertical asymptote: x 

x2 4  x  2  x  2 x  2

Vertical asymptote: x  2

1 3

Slant asymptote: y  x  2

Horizontal asymptote: y  0

y

y 10 3

8 6

1 −3

−2

−1

4 x 1

2

2

−1 −8

−2

−6

x

−4

4

−3

87. x2   y  82  25

88. x  42   y  7  4

y

x  42  y  7  4

14

Circle

12

Center: 0, 8

x  42    y  3

10 8

Parabola

6 4

Vertex: 4, 3

2 −8 −6 −4 −2

y

x 2

4

6

8

x

−2

2 −2

P   14

−4

Focus: 4, 3.25

−6

Directrix: y  2.75

−8 − 10

89. f x  2x1  5

90. f x  2x1  1

Horizontal asymptote: y  5 5

x f x

3

5.02

5.06

1

Horizontal asymptote: y  1 0

5.3

5.5

1 6

3 9

5

x

21

f x

2 3

y

1

0

1

2

2

 32

 54

8

y

14

2

12

x

−2

10 8 6

−4

4

−6

2 −6 −4 −2

−8

x 2

4

6

8 10

− 10

91. f x  3x  4

y 5 4 3 2 1

Horizontal asymptote: y  4 x

4

2

1

0

1

2

f x

3.99

3.89

3.67

3

1

5

− 6 − 5 − 4 − 3 − 2 −1 −2 −3 −5

x 2 3 4

9

4

6

8

Review Exercises for Chapter 3 92. f x  3x  4

317

y

Horizontal asymptote: y  4

5

x

2

1

0

1

2

f x

389

323

3

1

5

2 1 − 5 − 4 −3 −2 − 1

x 1 2 3 4 5

−2 −3 −4 −5

Review Exercises for Chapter 3 1.

f x  6.1x

2.

f x  30x

3. f x  20.5x

f 3   303  361.784

f 2.4  6.12.4  76.699 4. f x  1278x5

5.

f   20.5  0.337

f x  70.2x f  11   70.211 

f 1  127815  4.181

f x  145x

6.

f 0.8  1450.8  3.863

 1456.529 7. f x  4x

8. f x  4x

9. f x  4x

Intercept: 0, 1

Intercept: 0,1

Intercept: 0, 1

Horizontal asymptote: x-axis

Horizontal asymptote: y  0

Horizontal asymptote: x-axis

Increasing on:  , 

Decreasing on:  , 

Decreasing on:  , 

Matches graph (c).

Matches graph (d).

Matches graph (a).

10. f x  4x  1

12. f x  4x, gx  4x  3

11. f x  5x

Intercept: 0, 2

gx  5x1

Horizontal asymptote: y  1

Since gx  f x  1, the graph of g can be obtained by shifting the graph of f one unit to the right.

Increasing on:  , 

Because gx  f x  3, the graph of g can be obtained by shifting the graph of f three units downward.

Matches graph (b). 2 2 14. f x  3  , gx  8  3 

1 13. f x  2 

x

x

gx   12 

x2

Because gx  f x  8, the graph of g can be obtained by reflecting the graph of f in the x-axis and shifting the graph of f eight units upward.

Since gx  f x  2, the graph of g can be obtained by reflecting the graph of f about the x-axis and shifting f two units to the left. 15. f x  4x  4

y

Horizontal asymptote: y  4

8

x

1

0

1

2

3

f x

8

5

4.25

4.063

4.016 2 x −4

x

−2

2

4

318

Chapter 3

Exponential and Logarithmic Functions 17. f x  2.65x1

16. f x  4x  3

Horizontal asymptote: y  0

Horizontal asymptote: y  3 x

2

1

0

1

2

f x

3.063

3.25

4

7

19

x

2

1

0

1

2

f x

0.377

1

2.65

7.023

18.61

y

y −6

1 1

2

3

6

9

−3

x

−6 −5 −4 −3 −2 −1

x

−3

3

−6

−2 −3

−9

−4 −5

− 12

−6

− 15

−7 −8

19. f x  5x2  4

18. f x  2.65x1

Horizontal asymptote: y  4

Horizontal asymptote: y  0 x

3

1

0

1

3

x

1

0

1

2

3

f x

0.020

0.142

0.377

1

7.023

f x

4.008

4.04

4.2

5

9

y

y 8

5 4

6

3 2 2

1 −3

−2

x

−1

1

2

x

3

−4

−1

−2

2

1 21. f x  2 

x

20. f x  2x6  5 Horizontal asymptote: y  5

4

 3  2x  3

Horizontal asymptote: y  3

x

0

5

6

7

8

9

x

2

1

0

1

2

f x

4.984

4.5

4

3

1

3

f x

3.25

3.5

4

5

7

y

y 8

6 4

6 2 −2

x −2

2

4

6

10 2

−4 −6

x −4

−2

2

4

Review Exercises for Chapter 3 22. f x  18 

x2

5

y

Horizontal asymptote: y  5

2 x

x

3

2

1

0

2

f x

3

4

4.875

4.984

5

−4

2

4

−2 −4 −6

23.

3x2  19

24.

3x2  32 x  2  2

13 x2  81 13 x2  34 13 x2  13 4

x  4

e5x7  e15

25.

e82x  e3

26.

8  2x  3

5x  7  15

2x  11

5x  22 x

x  2  4

x

22 5

11 2

x  2 27. e8  2980.958

28. e58  1.868

29. e1.7  0.183

30. e0.278  1.320

32. hx  2  ex2

31. hx  ex2 x

2

1

0

1

2

x

2

1

0

1

2

hx

2.72

1.65

1

0.61

0.37

y

0.72

0.35

1

1.39

1.63

y

y 3

7 6 5 4

−4 −3

3

−1

x 1

2

3

4

−2

2

−3 −4 −3 −2 −1

−4

x 1

2

3

4

−5

33. f x  e x2

34. st  4e2t, t > 0

x

3

2

1

0

1

t

1 2

1

2

3

4

f x

0.37

1

2.72

7.39

20.09

y

0.07

0.54

1.47

2.05

2.43

y y 7 5

6

4 3 2

2

1 − 6 − 5 − 4 −3 − 2 − 1

x 1

1

2 t 1

2

3

4

5

319

320

Chapter 3



35. A  3500 1 

Exponential and Logarithmic Functions 0.065 n



10n

or A  3500e0.06510

n

1

2

4

12

365

Continuous Compounding

A

\$6569.98

\$6635.43

\$6669.46

\$6692.64

\$6704.00

\$6704.39



36. A  2000 1 

0.05 n



30n

or A  2000e0.0530

n

1

2

4

12

365

Continuous

A

\$8643.88

\$8799.58

\$8880.43

\$8935.49

\$8962.46

\$8963.38

37. Ft)  1  et3 (a) F 12  0.154

(b) F2  0.487

(c) F5  0.811

3 38. Vt  14,000 4

t

(a)

39. (a) A  50,000e0.087535  \$1,069,047.14

15,000

(b) The doubling time is

0

ln 2  7.9 years. 0.0875

10 0

(b) V2  14,00034   \$7875 2

(c) According to the model, the car depreciates most rapidly at the beginning. Yes, this is realistic. 40. Q  10012 

t14.4

(a) For t  0: Q  10012 

014.4

(c)

 100 grams

(b) For t  10: Q  10012 

1014.4

 61.79 grams

Mass of 241Pu (in grams)

Q

100 80 60 40 20 t 20

40

60

80 100

Time (in years)

41.

43  64

42.

log4 64  3 44.

e0  1 ln 1  0

2532  125 3 log25 125  2

45.

f x  log x f 1000  log 1000  log 103  3

43.

e0.8  2.2255 . . . ln 2.2255 . . .  0.8

46. log9 3  log9 912  21

Review Exercises for Chapter 3 48. f x  log4 x

47. gx  log2 x g

1 8



f

   log2 23  3

log2 18

1 4



log4 14

321

49. log4x  7  log4 14  1

x  7  14 x7

50. log83x  10  log8 5

51. lnx  9  ln 4

52. ln2x  1  ln11

x94

3x  10  5

2x  1  11

x  5

3x  15

2x  12 x6

x5 53. gx  log7 x ⇒ x  7y Domain: 0, 

3

gx

1 7

1

1

0

55. f x  log

7 1

x

1 −2

49

−1

x 1

2

3

2 1

50

x −1

x1

4

−1

1

−1

2

3

4

5

4

6

8

10

−2

x-intercept: 1, 0

−2

2

3

log5 x  0

2

Vertical asymptote: x  0

y

Domain: 0, 

4

x-intercept: 1, 0

x

54. gx  log5 x ⇒ 5y  x

y

−3

Vertical asymptote: x  0

3x  ⇒ 3x  10

y

⇒ x  310 y

Domain: 0, 

1 25

1 5

1

5

25

gx

2

1

0

1

2

56. f x  6  log x

1

x

0.03

0.3

3

30

f x

2

1

0

1

8 6

log x  6

2

−1

10

6  log x  0

3

Vertical asymptote: x0

y

Domain: 0, 

y

x-intercept: 3, 0

x

3

4

2

x  106

x 2

4

5

−2

x  0.000001

−1 −2

x 2 −2

x-intercept: 0.000001, 0

−3

Vertical asymptote: x  0

57. f x  4  logx  5

x

Domain: 5, 

4

3

2

x

1

2

4

6

8

10

f x

6

6.3

6.6

6.8

6.9

7

1

0

y

1 7

f x

x-intercept: 9995, 0

4

3.70

3.52

3.40

3.30

3.22

6 5 4

Since 4  logx  5  0 ⇒ logx  5  4

3 2

x  5  104 x  10  5  9995. 4

Vertical asymptote: x  5

1 −6

−4 −3 −2 −1

x 1

2

322

Chapter 3

Exponential and Logarithmic Functions

58. f x  logx  3  1

y

Domain: 3,  logx  3  1  0 logx  3  1 x3

x

4

5

6

7

8

f x

1

1.3

1.5

1.6

1.7

5 4 3 2 1 x −1

101

1 2

4 5 6 7 8 9

−2 −3 −4 −5

x  3.1 x-intercept: 3.1, 0 Vertical asymptote: x  3 59. ln 22.6  3.118

60. ln 0.98  0.020

61. ln e12  12

62. ln e7  7

63. ln7  5  2.034

64. ln

65. f x  ln x  3 6 5

x-intercept: ln x  3  0

4

ln x  3

4

lnx  3  0

2

x  3  e0

2

x 2

x4

1

e3, 0

y

Domain: 3, 

3

x  e3



66. f x  lnx  3

y

Domain: 0, 

 83   1.530

x

−1

1

2

3

4

5

Vertical asymptote: x  0

4

x-intercept: 4, 0

−4

Vertical asymptote: x  3

1

2

3

1 2

1 4

x

3.5

4

4.5

5

5.5

f x

3

3.69

4.10

2.31

1.61

y

0.69

0

0.41

0.69

0.92



67. hx  lnx2  2 ln x

Domain:  , 0  0,

4

x-intercepts: ± 1, 0

2

3

Domain: 0, 

3

1 4

1 −4 −3 −2 −1

y

68. f x  14 ln x

y

2

3

1 x

ln x  0

x 1

2

ln x  0

4

1

−3

3

4

5

−2

x1

−4

2

−1

x  e0

−3

x-intercept: 1, 0

69.

x

± 0.5

±1

±2

y

1.39

0

1.39 2.20

±3

h  116 loga  40  176 h55  116 log55  40  176  53.4 inches

8

−2

x

Vertical asymptote: x  0

6

±4

Vertical asymptote: x  0

2.77

70. s  25 

13 ln1012 ln 3

 27.16 miles

x

1 2

1

3 2

2

5 2

3

y

0.17

0

0.10

0.17

0.23

0.27

71. log4 9  log4 9 

log 9  1.585 log 4 ln 9  1.585 ln 4

6

Review Exercises for Chapter 3

72. log12 200  log12 200 

log 200  2.132 log 12

73. log12 5 

ln 200  2.132 ln 12

75. log 18  log2

log12 5 

log 5  2.322 log12

 32

76. log2

ln 0.28  1.159 ln 3

log3 0.28 

1  log2 1  log2 12  0  log22 12

 log 2  2 log 3

 2 log2 22  log2 3  2 

 1.255

77. ln 20  ln22

log 0.28  1.159 log 3

74. log3 0.28 

ln 5  2.322 ln12

 3

log 3 log 2

 3.585

 5

78. ln 3e4  ln 3  ln e4

79. log5 5x2  log5 5  log5 x2

 ln 3  4

 2 ln 2  ln 5  2.996

 1  2 log5 x

 2.90

80. log10 7x 4  log 7  log x 4

81. log3

 log 7  4 log x

6 3

x

3 x  log3 6  log3 

 log33

 1  log3 2 

86. ln

 log7 x12  log7 4

1 log3 x 3



1 log7 x  log7 4 2

1 log3 x 3

y 4 1

2

 2 ln

y 4 1

 lnx  3  ln x  ln y

 2 ln y  1  2 ln 4

 lnx  3  ln x  ln y

 2 ln y  1  ln 16, y > 1 88. log6 y  2 log6 z  log6 y  log6 z2

87. log2 5  log2 x  log2 5x

 log6

91.

 log7 x  log7 4

 ln 3  ln x  2 ln y

x xy 3  lnx  3  ln xy

89. ln x 

4

84. ln 3xy2  ln 3  ln x  ln y 2

 2 ln x  2 ln y  ln z

85. ln

x

 2  log3 x13

 log3 3  log3 2 

83. ln x2y 2z  ln x2  ln y 2  ln z

82. log7

 

x 1 4 y  ln ln y  ln x  ln  4 4  y

1 3 x  4  log y7 log8x  4  7 log8 y  log8  8 3 3  log8 y7  x  4

323

y z2

90. 3 ln x  2 lnx  1  ln x3  lnx  12  ln x3x  12

92. 2 log x  5 logx  6  log x2  logx  65  log

x2 x  65

 log

1 x2x  65

324

93.

Chapter 3

Exponential and Logarithmic Functions

1 ln2x  1  2 lnx  1  ln2x  1  lnx  12 2  ln

2x  1

x  12

94. 5 lnx  2  lnx  2  3 ln x  lnx  25  lnx  2  ln x3  lnx  25  lnx  2  ln x3

 lnx  25  ln x3x  2  ln

95. t  50 log

x  25 x3x  2

18,000 18,000  h

(a) Domain: 0 ≤ h < 18,000 (b)

(c) As the plane approaches its absolute ceiling, it climbs at a slower rate, so the time required increases.

100

(d) 50 log

0

18,000  5.46 minutes 18,000  4000

20,000 0

Vertical asymptote: h  18,000 96. Using a calculator gives s  84.66  11 ln t.

ex  6

100. ln

ex

1 98. 6x  216

97. 8x  512 8x  83

6x  63

x3

x  3

101. log4 x  2

 ln 6

103. ln x  4

61

x  e4

x  4  16

6log6 x



x  16

x  ln 6  1.792 ex  12

105.

x  e3  0.0498

106.

ln ex  ln12

e3x2  40 ln e3x2  ln 40 3x  2  ln 40 x

ln 40  2  0.563 3

107. e4x  ex

109. 2x  13  35 2x  22 x  log2 22 

4x  x 2  3

3x  ln 25 x

14e3x2  560

2 3

e3x  25 ln e3x  ln 25

x  ln 12  2.485

108.

x  ln 3

102. log6 x  1

2

104. ln x  3

99. ex  3

log 22 ln 22 or log 2 ln 2

x  4.459

0  x 2  4x  3 0  x  1x  3

ln 25  1.073 3

x  1 or x  3

110. 6x  28  8 6x  20 log6 6x  log6 20 x  log6 20 x

ln 20  1.672 ln 6

Review Exercises for Chapter 3 111. 45x  68

112. 212x  190

5x  17

12x  95

ln 5x  ln 17

ln 12x  ln 95

x ln 5  ln 17

x ln 12  ln 95

ln 17  1.760 ln 5

x

x

113. e2x  7e x  10  0 ex  2

ex  2ex  4  0 ex  5

or

ln e x  ln 2

ex  2

ln e x  ln 5

x  ln 2  0.693

x  ln 5  1.609

115. 20.6x  3x  0

3x 

e8.2

x

x  1.386

 x. −12

6 −3

12

Graph y1  4e1.2x and y2  9. −6

The graphs intersect at x  0.676.

18

−6

6 −2

−2

120. ln 5x  7.2 5x  x

e8.2  1213.650 3

9

40.2x

118. 4e1.2x  9

16

Graph y1  25e0.3x and y2  12.



x  0.693

The x-intercepts are at x  7.038 and at x  1.527.

−10

119. ln 3x  8.2

x  ln 4

10

The x-intercepts are at x  0.392 and at x  7.480.

117. 25e0.3x  12

ex  4

x  ln 2

Graph y1  −10

The graphs intersect at x  2.447.

or

116. 40.2x  x  0

10

Graph y1  20.6x  3x.

e8.2

ln 95  1.833 ln 12

e2x  6ex  8  0

114.

e x  2e x  5  0

eln 3x

325

121. 2 ln 4x  15

e7.2

ln 4x 

7.2

e  267.886 5

15 2

eln 4x  e7.5 4x  e7.5 1 x  e7.5  452.011 4

122. 4 ln 3x  15 ln 3x 

123. ln x  ln 3  2

15 4

x

e154 3

 14.174

lnx  8  3

x 2 3

1 lnx  8  3 2

elnx3  e2

lnx  8  6

x  e2 3

x  8  e6

ln

3x  e154

124.

x  e6  8  395.429

x  3e  22.167 2

326 125.

Chapter 3

Exponential and Logarithmic Functions

lnx  1  2

126. ln x  ln 5  4

1 lnx  1  2 2

ln

lnx  1  4 elnx1



x 4 5 x  e4 5

e4

x  5e4  272.991

x  1  e4 x  e4  1  53.598 log8x  1  log8x  2  log8x  2

127.

log8x  1  log8 x1

128. log6x  2  log6 x  log6x  5

x2

x  2

log6

x x 2  log x  5 6

x2 x5 x

x2 x2

x  2  x2  5x

x  1x  2  x  2

0  x2  4x  2

x2  x  2  x  2

x  2 ± 6, Quadratic Formula

x2  0

Only x  2  6  0.449 is a valid solution.

x0 Since x  0 is not in the domain of log8x  1 or of log8x  2, it is an extraneous solution. The equation has no solution. 129. log1  x  1

130. logx  4  2

1  x  10

x  4  102

1

1 1  10 x

x  100  4

x  0.900

x  104

131. 2 lnx  3  3x  8

132. 6 logx 2  1  x  0

Graph y1  2 lnx  3  3x and y2  8.

Graph y1  6 logx 2  1  x. 12

10

(1.64, 8) −8

−9

16

9

−4

−2

The graphs intersect at approximately 1.643, 8. The solution of the equation is x  1.643. 133. 4 lnx  5  x  10 Graph y1  4 lnx  5  x and y2  10. 11

−6

12 −1

The graphs do not intersect. The equation has no solution.

The x-intercepts are at x  0, x  0.416, and x  13.627.

Review Exercises for Chapter 3 135. 37550  7550e0.0725t

134. x  2 logx  4  0

3  e0.0725t

Let y1  x  2 logx  4.

ln 3  ln e0.0725t

12

ln 3  0.0725t −8

t

16 −4

ln 3  15.2 years 0.0725

The x-intercepts are at x  3.990 and x  1.477. 136.

S  93 logd  65

137. y  3e2x3

283  93 logd  65

Exponential decay model

218  93 logd

Matches graph (e).

logd 

218 93

d  1021893  220.8 miles 139. y  lnx  3

138. y  4e2x3 Exponential growth model

Logarithmic model

Matches graph (b).

Vertical asymptote: x  3 Graph includes 2, 0 Matches graph (f).

140. y  7  logx  3

141. y  2ex4 3 2

142. y 

Logarithmic model

Gaussian model

Vertical asymptote: x  3

Matches graph (a).

Logistics growth model Matches graph (c).

Matches graph (d). 143.

y  aebx

144.

2  aeb0 ⇒ a  2 3

2eb4

1.5  e4b ln 1.5  4b

⇒ b  0.1014

Thus, y  2e0.1014x.

6 1  2e2x

y  aebx 1 1  aeb0 ⇒ a  2 2 1 5  eb5 2 10  e5b ln 10  5b ln 10 b 5 b  0.4605 1 y  e0.4605x 2

327

328

Chapter 3

Exponential and Logarithmic Functions

P  3499e0.0135t

145.

4.5 million  4500 thousand 4500 

ln

y  Cekt

146.

1 C  Ce250,000k 2

3499e0.0135t

4500  e0.0135t 3499

ln

1  ln e250,000k 2

 0.0135t 4500 3499 

ln

1  250,000k 2

t

ln45003499  18.6 years 0.0135

k

ln12 250,000

When t  5000, we have

According to this model, the population of South Carolina will reach 4.5 million during the year 2008.

y  Ce ln12250,000 5000  0.986C  98.6%C. After 5000 years, approximately 98.6% of the radioactive uranium II will remain.

147. (a) 20,000  10,000er5 2  e5r

2

40 ≤ x ≤ 100

1400  2000e3k

ln 2  5r

(a) Graph y1  0.0499ex71 128. 2

7  e3k 10

ln 2 r 5 r  0.138629

3k  ln

 13.8629% (b) A 

149. y  0.0499ex71 128,

148. N0  2000 and N3  1400 so N  2000ekt and:

k

10,000e0.138629

 \$11,486.98

0.05

107 

ln710  0.11889 3

40

100 0

(b) The average test score is 71.

The population one year ago: N4  2000e0.118894  1243 bats

150. N 

157 1  5.4e0.12t

(a) When N  50: 50 

(b) When N  75: 157 1  5.4e0.12t

75 

1  5.4e0.12t 

157 50

1  5.4e0.12t 

5.4e0.12t 

107 50

5.4e0.12t 

e0.12t 

107 270

e0.12t 

0.12t  ln t

107 270

ln107270  7.7 weeks 0.12

157 1  5.4e0.12t 157 75 82 75 82 405

0.12t  ln t

82 405

ln82405  13.3 weeks 0.12

Problem Solving for Chapter 3

  10 log

151.

125  10 log 12.5  log 1012.5 

10  I

152. R  log I since I0  1.

16



I 1016

329

(a) log I  8.4



I  108.4  251,188,643 (b) log I  6.85

10  I

16

I  106.85  7,079,458 (c) log I  9.1

I 1016

I  109.1  1,258,925,412

I  103.5 wattcm2

154. False. ln x  ln y  lnxy  lnx  y

153. True. By the inverse properties, logb b2x  2x.

155. Since graphs (b) and (d) represent exponential decay, b and d are negative. Since graph (a) and (c) represent exponential growth, a and c are positive.

Problem Solving for Chapter 3 1. y  ax 0.5x

7

y2  1.2x

5

y3  2.0x

3

y1 

2. y1  ex

y

y2 

y3

6

y4  x

1

2

3

y4 0

6 0



y5  x

y1 −4 −3 −2 −1

y2 y5

y4  x

y2

2

y1

y3

y3  x3

y4

4

24

x2

x

The function that increases at the fastest rate for “large” values of x is y1  ex. (Note: One of the intersection points of y  ex and y  x3 is approximately 4.536, 93 and past this point ex > x3. This is not shown on the graph above.)

4

The curves y  0.5x and y  1.2x cross the line y  x. From checking the graphs it appears that y  x will cross y  ax for 0 ≤ a ≤ 1.44. 3. The exponential function, y  ex, increases at a faster rate than the polynomial function y  xn.

4. It usually implies rapid growth.

5. (a) f u  v  auv

6. f x 2  g x 2 

e

 ex 2



e

 2  e2x e2x  2  e2x  4 4



4 4

 au

 av

 f u  f v (b) f 2x  a2x  ax2

(b)

6

y = ex

2x

  e

x

2

 ex 2

 

1

 f x 2 7. (a)

x

(c)

6

6

y = ex

y1

y = ex

y2 −6

6 −2

−6

6 −2

−6

6

y3 −2



2



330

Chapter 3

Exponential and Logarithmic Functions

x x2 x3 x4    1! 2! 3! 4!

8. y4  1 

f x  e x  ex

9.

6

y4

y = ex

−6

y  e x  ex x

2

ey



3

ey

1

e2y  1 x ey

6 −2

x

− 4 − 3 − 2 −1

xe y  e2y  1

As more terms are added, the polynomial approaches ex.

1

2

3

4

−4

e 2y  xe y  10

x x2 x3 x4 x5     . . . 1! 2! 3! 4! 5!

ex  1 

y

4

ey 

x ± x2  4 2

Choosing the positive quantity for e y we have



y  ln

ax  1 , a > 0, a  1 ax  1

10. f x 

x







x  x2  4 x  x2  4 . Thus, f 1x  ln . 2 2

11. Answer (c). y  61  ex 2 2

The graph passes through 0, 0 and neither (a) nor (b) pass through the origin. Also, the graph has y-axis symmetry and a horizontal asymptote at y  6.

ay  1 ay  1

xay  1  ay  1 xay  ay  x  1 ayx  1  x  1 x1 x1 ln

x1 y  loga  x1





xx  11 ln a

 f 1x

12. (a) The steeper curve represents the investment earning compound interest, because compound interest earns more than simple interest. With simple interest there is no compounding so the growth is linear. (b) Compound interest formula: A  5001  0.07 1 

1t

 5001.07t

Simple interest formula: A  Prt  P  5000.07t  500

A Compounded Interest

Growth of investment (in dollars)

ay 

(c) One should choose compound interest since the earnings would be higher.

13. y1  c1

12



1 c1 2

tk1

tk1

and y2  c2



1  c2 2



c1 1  c2 2

12

tk2

ln c1  ln c2  t t

1

k1  k1  ln12 2

1000

Simple Interest t 5 10 15 20 25 30

Time (in years)

B0  500

tk2 tk1

2

2000

200  500ak2

1 2

3000

14. B  B0akt through 0, 500 and 2, 200

tk2

cc   kt  kt  ln12

ln

4000

1

ln c1  ln c2 1k2  1k1 ln12

2  a2k 5 loga

25  2k 

1 2 loga k 2 5 B  500a 12 loga 25 t  500 a log a 25 t2  500

25

t2

Problem Solving for Chapter 3 15. (a) y  252.6061.0310t

16. Let loga x  m and logab x  n. Then x  am and x  abn.

(b) y  400.88t  1464.6t  291,782 2

(c)

am 

2,900,000

y2

amn 

a b

amn1 

1 b

y1 0 200,000

ab

n

85

(d) Both models appear to be “good fits” for the data, but neither would be reliable to predict the population of the United States in 2010. The exponential model approaches infinity rapidly.

loga

1 m  1 b n

1  loga

1 m  b n

1  loga

1 loga x  b logab x

ln x2  ln x2

17.

ln x2  2 ln x  0 ln xln x  2  0 ln x  0 or ln x  2 x  1 or

x  e2

18. y  ln x y1  x  1 y2  x  1  12x  12 y3  x  1  12x  12  13x  13 (a)

(b)

4

y1 −3

(c)

4

y = ln x

4

y = ln x 9

−3

−4

y3 9

y2

y = ln x

−3

9

−4

−4

19. y 4  x  1  12x  12  13x  13  14x  14

4

y = ln x

The pattern implies that ln x  x  1  12x  12  13x  13  14x  14  . . . .

−3

9

y4 −4

20. y  abx

y  axb

ln y  lnabx

ln y  lnax b

ln y  ln a  ln bx

ln y  ln a  ln x b

ln y  ln a  x ln b

ln y  ln a  b ln x

ln y  ln bx  ln a

ln y  b ln x  ln a

Slope: m  ln b

Slope: m  b

y-intercept: 0, ln a

y-intercept: 0, ln a

21. y  80.4  11 ln x 30

100

1500 0

y300  80.4  11 ln 300  17.7 ft3min

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332

Chapter 3

Exponential and Logarithmic Functions

22. (a)

450  15 cubic feet per minute 30 15  80.4  11 ln x

(b)

11 ln x  65.4 ln x  x

V  xh

x  382

e65.411

9

0

Let x  floor space in square feet and h  30 feet. 11,460  x30

65.4 11

x  382 cubic feet of air space per child. 23. (a)

(c) Total air space required: 38230  11,460 cubic feet

If the ceiling height is 30 feet, the minimum number of square feet of floor space required is 382 square feet.

24. (a)

9

36

0

9

0

0

(b) The data could best be modeled by a logarithmic model.

(b) The data could best be modeled by an exponential model.

(c) The shape of the curve looks much more logarithmic than linear or exponential.

(c) The data scatter plot looks exponential.

(d) y  2.1518  2.7044 ln x

(d) y  3.1141.341x 36

9

0 0

9 0

9 0

(e) The model graph hits every point of the scatter plot. (e) The model is a good fit to the actual data.

25. (a)

26. (a)

9

0

9

10

0

0

(b) The data could best be modeled by a linear model. (c) The shape of the curve looks much more linear than exponential or logarithmic. (d) y  0.7884x  8.2566

(b) The data could best be modeled by a logarithmic model. (c) The data scatter plot looks logarithmic. (d) y  5.099  1.92 lnx

9

0

9 0

10

9 0

(e) The model is a good fit to the actual data.

0

9 0

(e) The model graph hits every point of the scatter plot.

Practice Test for Chapter 3

Chapter 3

Practice Test

1. Solve for x: x35  8. 1

2. Solve for x: 3x1  81. 3. Graph f x  2x. 4. Graph gx  ex  1. 5. If \$5000 is invested at 9% interest, find the amount after three years if the interest is compounded (a) monthly.

(b) quarterly.

(c) continuously.

1 6. Write the equation in logarithmic form: 72  49. 1

7. Solve for x: x  4  log2 64. 4 825. 8. Given logb 2  0.3562 and logb 5  0.8271, evaluate logb 

1

9. Write 5 ln x  2 ln y  6 ln z as a single logarithm. 10. Using your calculator and the change of base formula, evaluate log9 28. 11. Use your calculator to solve for N: log10 N  0.6646 12. Graph y  log4 x. 13. Determine the domain of f x  log3x2  9. 14. Graph y  lnx  2.

15. True or false:

ln x  lnx  y ln y

16. Solve for x: 5x  41 1 17. Solve for x: x  x2  log5 25

18. Solve for x: log2 x  log2x  3  2

19. Solve for x:

ex  ex 4 3

20. Six thousand dollars is deposited into a fund at an annual interest rate of 13%. Find the time required for the investment to double if the interest is compounded continuously.

333