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 3.45.6 946.852
2. f x 2.3x 2.332 3.488
3. f 5 0.006
4. f x 23 23
5. gx 50002x 500021.5
6. f x 2001.212x
5x
50.3
0.544
2001.212 24
1767.767
1.274 1025 7. f x 2x
9. f x 2x
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 2x2 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 6x
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 2x1
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 4x3 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, gx 3x4 Because gx 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, gx 5 2x Because gx 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, gx 4x 1 Because gx f x 1, the graph of g can be obtained by shifting the graph of f one unit upward. 20. f x 10x, gx 10x3 Because gx 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 , gx 72
x6
x
gx 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 3x
2
25. f x 3x2 1
3
3
26. y 4x1 2 3
4
−6 −3
267
22. f x 0.3x, gx 0.3x 5
Because gx 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 2x
Exponential Functions and Their Graphs
−3
3
3
3 −1
−1
−1
5
−3
0
3 27. f 4 e34 0.472
28. f x ex e3.2 24.533
29. f 10 2e510 3.857 1022
30. f x 1.5e12x
31. f 6 5000e0.066 7166.647
32. f x 250e0.05x 250e0.0520 679.570
1.5e120 1.956 1052 33. f x e x
34. f x ex
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 2e0.5x
35. f x 3e x4 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 x2 4
38. f x 2 ex5
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. st 2e0.12t 22
−4
−7
8
5 −10
−2
−1
42. st 3e0.2t
44. hx ex2 4
4
− 16
17
−3
−2
−2
3
46.
3x1 33
4 0
0
3x1 27
23 0
43. gx 1 ex
20
2x3 16
47.
2x3 24
1 2x2 32
2x2 25
x13
x34
x 2 5
x2
x7
x 3
15 x1 125 15 x1 53 15 x1 15 3
49.
e3x2 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 3x 1 0
x 3x 2 0
x 3 or x 1
e2x1 e4
3x 2 3 x
x 1 3
51.
8
6
7
48.
2
40. y 1.085x
39. y 1.085x
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 2500e0.02510 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 1000e0.0410
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 2500e0.0320 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 1000e0.0640
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 25,000e0.087525
62. A 5000e0.07550
$222,822.57
64. p 5000 1 (a)
63. C10 23.951.0410 $35.45
$212,605.41
4 4 e0.002x
65. Vt 100e4.6052t (a) V1 10,000.298 computers
1200
(b) V1.5 10,004.472 computers (c) V2 1,000,059.63 computers 0
2000 0
(b) When x 500:
p 5000 1
4 $421.12 4 e0.002500
(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 2512
t1599
66. (a) P 152.26e0.0039t
(a) Q0 25 grams
Since the growth rate is negative, 0.0039 0.39%, the population is decreasing.
(b) Q1000 16.21 grams
(b) In 1998, t 8 and the population is given by P8 152.26e0.00398 147.58 million.
(c)
30
In 2000, t 10 and the population is given by P10 152.26e0.003910 146.44 million. 0
(c) In 2010, t 20 and the population is given by P20 152.26e0.003920 140.84 million. t5715
(a) When t 0: Q 1012
05715
101 10 grams (b) When t 2000: Q 102
1 20005715
7.85 grams
(c)
Q
Mass of 14C (in grams)
68. Q 1012
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 7e0.069x
69. y (a)
Section 3.1
(b)
110
0
120 0
70. (a)
y (d)
100 63.14%. 1 7e0.06936
2 100 100 when 3 1 7e0.069x x 38 masses.
(b) p 107,428e0.150h
P
Atmospheric pressure (in pascals)
(c) When x 36:
120,000
107,428e0.1508
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 3x2
74. gx 22x6 22x 26
3x32 1
6422x
2
6422x
3
3x
1 3x 9
644x hx
hx Thus, f x gx, but f x hx. 75. f x 164x
e is an irrational number.
and
f x 164x
Thus, gx hx but gx f x.
76. f x 5x 3
5x
424x
1622x
gx 53x 53
42x
1622x
hx 5x3 5x 53
hx
Thus, none are equal.
2x
14 4
1 x2
gx Thus, f x gx hx.
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) gx x23x
78. (a) f x x2ex
6
5
−2
−2
7
10 −2
−1
Decreasing: , 0, 2,
Decreasing: 1.44,
Increasing: 0, 2
Increasing: , 1.44
Relative maximum: 2, 4e2
Relative maximum: 1.44, 4.25
Relative minimum: 0, 0
79. f x 1
0.5 x
and gx e x
0.5
80. The functions (c) 3x and (d) 2x are exponential.
(Horizontal line)
4
f g −3
3 0
As x → , f x → gx. As x → , f x → gx.
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 9x
y 12
Vertical asymptote: x 9
9 6
Horizontal asymptote: y 0 x
11
10
f x
1
2
x2 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
85. Answers will vary.
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 ⇒ 72 491
1 1 4. log 1000 3 ⇒ 103 1000
2 5. log32 4 5 ⇒ 3225 4
6. log16 8 34 ⇒ 1634 8
1 7. log36 6 2 ⇒ 36 12 6
8. log8 4 23 ⇒ 823 4
9. 53 125 ⇒ log5 125 3
10. 82 64 ⇒ log8 64 2
1 11. 8114 3 ⇒ log81 3 4
3 12. 932 27 ⇒ log9 27 2
273
274
Chapter 3
Exponential and Logarithmic Functions
1 1 13. 62 36 ⇒ log6 36 2
1 1 14. 43 64 ⇒ log4 64 3
15. 70 1 ⇒ log7 1 0
16. 103 0.001 ⇒ log10 0.001 3
17. f x log2 x
18. f x log16 x f 4 log16 4 12 since 1612 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. gx loga x
f 10 log 10 1 since 101 10
ga2 loga a2 2 by the Inverse Property
22. gx logb x g
b3
logb
23. f x log x b3
f
3 since
4 5
24. f x log x
log 0.097
1 f 500 log 5001 2.699
4 5
b3 b3 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. gx 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
log4x 3 0
4 2
40 x 3
−6
9x The x-intercept is 9, 0.
x 2
1x3
−2
4x
−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 ⇒
32y
x
3
Domain: x 3 > 0 ⇒ x > 3
6
x-intercept:
2
34. hx log4x 3
y
Domain: 0,
1 −1
x
x
x
−1
y log4x 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 log6x 2
Logarithmic Functions and Their Graphs
36. y log5x 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 log6x 2
4
x-intercept:
2
log5x 1 4 0
0 log6x 2
6 −2
1x2
−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:
x1
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 ⇒ x5 5
−4
The x-intercept is 5, 0. Vertical asymptote:
1 625
x
The domain is 0, .
log
2
y log5x 1 4 ⇒ 5y4 1 x
6y 2 x
f x
3
Vertical asymptote: x 1 0 ⇒ x 1
y log6x 2
1
4
626
y log6x 2
4
5
The x-intercept is 625, 0.
Vertical asymptote: x 2 0 ⇒ x 2
x
6
54 x 1
1 x
156
y
log5x 1 4
x
60 x 2
275
x 0 ⇒ x0 5
The vertical asymptote is the y-axis. 38. y logx Domain: x > 0 ⇒ x < 0 The domain is , 0.
x
1 100
10
1
1
10
y
2
1
0
1
x-intercept: logx 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 logx ⇒ 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 gx upward two units.
f x reflects gx in the x-axis.
41. f x log3x 2
42. f x log3x 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 gx about the x-axis and shifting the graph two units to the left.
f x shifts gx one unit to the right.
43. f x log31 x log3 x 1
44. f x log3x
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 gx about the y-axis and shifting the graph one unit to the right.
f x reflects gx in the x-axis then reflects that graph in the y-axis.
45. ln 12 0.693 . . . ⇒ e0.693 . . . 12
46. ln 25 0.916 . . . ⇒ e0.916 . . . 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. e12 1.6487 . . . ⇒ ln 1.6487 . . . 12
1 56. e13 1.3956 . . . ⇒ ln 1.3956 . . . 3
57. e0.5 0.6065 . . . ⇒ ln 0.6065 . . . 0.5
58. e4.1 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. gx 2 ln x g0.75 2 ln 0.75 0.575
f 0.32 3 ln 0.32 3.418 64. gx ln x g12 ln 12 0.693
. .
679
Section 3.2 65. gx ln x
Logarithmic Functions and Their Graphs
66. gx ln x
ge3 ln e3 3 by the Inverse Property
ge2 ln e2 2
67. gx ln x
68. gx ln x
ge23 ln e23
23
ge52 ln e52 52
by the Inverse Property
69. f x lnx 1
70. hx lnx 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 lnx 1
1
e0 x 1
−1
2x
6
lnx 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 lnx 1 ⇒ ey 1 x x
0.39
0
1.72
6.39
19.09
y
12
0
1
2
3
72. f x ln3 x
71. gx lnx 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 lnx
−3
−2
y 3 2
ln3 x 0
x
−1
1
e0 3 x
e0 x 1 x
−2
−1
13x
−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
gx
0.69
0
0.69
1.10
y ln3 x ⇒ 3 ey x
74. f x logx 1
73. y1 logx 1
x
2.95
2.86
2.63
2
0.28
y
3
2
1
0
1
75. y1 lnx 1 3
2
2
5
2
−1 −2
2x
Vertical asymptote: x 0 ⇒ x 0
−2
6
0x
Vertical asymptote: x 1 0 ⇒ x 1 1.5
x
−2
1x1
−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 lnx 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. log2x 3 log2 9
79. log2x 1 log2 4
x39
x14
x 12
x3 81. log2x 1 log 15
82. log5x 3 log 12
2x 1 15
5x 3 12
x7
5x 9 9
x5 83. lnx 2 ln 6
84. lnx 4 ln 2
x26
x42
x4
x6
85. lnx 2 2 ln 23
lnx 2 x ln 6
86.
x 2 2 23
x2 x 6
x 2 25
x2 x 6 0
x ±5
x 3x 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: 1100.651230 $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
1254.681220 $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 logt 1, 0 ≤ t ≤ 12 (a)
Logarithmic Functions and Their Graphs
90. 10 log
10I 12
100
(a) 10 log
0
101 10 log10
(b) 10 log
12
279
0
120 decibels
12
12 2
1010 10 log10
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 gx about the line y x will determine the graph of f x. 93. f x 3x, gx log3 x
94. f x 5x, gx log5 x
95 . f x ex, gx 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, gx
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, gx x
y
98. f x
1
−1
96. f x 10x, gx 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
10 31 2 1 1 and m2 . 21 82 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 2y 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) hx lnx2 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 g2 f 2 g2
104. f x gx 3x 2 x3 1
32 2 23 1
3x 2 x3 1
87
3x x3 3
15
Therefore,
f g1 31 13 3 3 1 3 1.
105. fg6 f 6g6 36 2 63 1 20215
106.
f x 3x 2 3 gx x 1 f 3 02 Therefore, 0 3 2. g 0 1
4300 107. f g7 f g7
108. g f (x g f x g3x 2 3x 23 1
f 73 1
Therefore,
f 342
g f 3 3 3 23 1
3342 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) logauv loga u loga v (c) loga
v log u
a
lnuv 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. logauv 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) log13 x (b) log13 x
7. (a) log2.6 x (b) log2.6 x
10. log7 4
log x log13 ln x ln13 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) log15 x
log310 3 log x 10
6. (a) logx
3 ln310 10 ln x
(b) log7.1 x
log 5 ln 5 1.161 log14 ln14
3. (a) log15 x
ln x ln 3
8. (a) log7.1 x
log 4 ln 4 0.712 log 7 ln 7
12. log14 5
log x log 3
(b) logx
log x log 7.1
log x log15 ln x ln15
3 log34 4 log x 3 ln34 4 ln x
9. log3 7
log 7 ln 7 1.771 log 3 ln 3
ln x ln 7.1
11. log12 4
log 4 ln 4 2.000 log12 ln12
13. log90.4
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. log242 34 log2 42 log2 34
1 1 19. log5 250 log5125
12
2 log2 4 4 log2 3
1 log5 125 log5 12
2 log2
log5
22
4 log2 3
53
log5
9 3 20. log 300 log 100
log 3 log 100 21
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. ln5e6 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 53 3 log5 5 31 3
4 8 1 log 23 3 log 2 3 1 3 25. log2 4 2 4 2 4 4
3 6 log 613 1 log 6 1 1 1 26. log6 6 3 6 3 3
27. log4 161.2 1.2log4 16 1.2 log4 42 1.22 2.4
28. log3 810.2 0.2 log3 81
29. log39 is undefined. 9 is not in the domain of log3 x.
0.2 log3 3
4
0.24 0.8 30. log216 is undefined because 16 is not in the domain of log2 x.
31. ln e4.5 4.5
32. 3 ln e4 34 ln e 121 12
33. ln
1 e
4 e3 ln e34 34. ln
ln 1 lne 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 412 log4 452 1 2
5 2
log4 4 log4 4 121 521 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 z3 3 log6 z
45. lnz ln z12
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 t13 1 ln t 46. ln 3
ln x ln y 2 ln z 49. ln zz 12 ln z lnz 12
50. ln
ln z 2 lnz 1, z > 1
log 4 2 log x log y
x2 1 lnx2 1 ln x3 x3 lnx 1x 1 ln x3
lnx 1 lnx 1 3 ln x
51. log2
a 1
9
log2a 1 log2 9
6
52. ln
x2 1
ln 6 lnx2 112
1 log2a 1 log2 32 2
ln 6
1 log2a 1 2 log2 3, a > 1 2
xy 31 ln yx
xy
2
3
53. ln
ln 6 lnx2 1
54. ln
3
ln
1 ln x ln y 3 1 1 ln x ln y 3 3
y x2
3
12
1 lnx2 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
4y
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 x3x2 3
1 3 4 ln x
lnx2 3
60. lnx2x 2 lnx2x 2 12 lnxx 212
14 3 ln x lnx2 3
ln x lnx 212
34 ln x 14 lnx2 3
ln x 12 lnx 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 log2x 4 log2x 42
66.
1 4 5x log3 5x log35x14 log3 4
ln
1 16x 4
70. 2 ln 8 5 lnz 4 ln 82 lnz 45 ln 64 lnz 45
x x 13
ln 64z 45
71. log x 2 log y 3 log z log x log y2 log z3 log
2 log7z 2 log7z 223 3
68. 4 log6 2x log62x4 log6
69. ln x 3 lnx 1 ln x lnx 13
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 4lnx 2 lnx 2 ln x 4 lnx 2x 2 ln x 4 lnx2 4 ln x lnx2 44 ln
x x2 44
74. 4ln z lnz 5 2 lnz 5 4ln zz 5 lnz 52 lnzz 5 4 lnz 52 ln
75.
z y
z4z 54 z 52
1 1 2 lnx 3 ln x lnx2 1 lnx 32 ln x lnx2 1 3 3 1 ln xx 32 lnx2 1 3
1 xx 32 ln 2 3 x 1
xxx 31
ln
2
3
2
76. 23 ln x lnx 1 lnx 1 2ln x3 lnx 1 lnx 1 2ln x3 lnx 1 lnx 1 2ln x3 lnx 1x 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 42 log 8 y 1 3 3
1 log 8 y y 42 log 8 y 1 3
3 log 8 y y 42 log 8 y 1
log 8
3 y y 42
y1
78. 12log4x 1 2 log4x 1 6 log4 x 12log4x 1 log4x 12 log4 x6 12log4x 1x 12 log4 x6 log4x 1x 1 log4 x6 log4x6x 1x 1
79. log2
32 log2 32 log2 32 log2 4 4 log2 4
The second and third expressions are equal by Property 2.
80. log770
1 1 log7 70 log7 7 log7 10 2 2
81. 10 log
10I 12
1 1 log7 10 2
10log I log 1012
1 1 log7 10 2 2
120 10 log I
10log I 12
When I 106 :
1 log7 10 by Property 1 and Property 3 2
120 10 log 106 120 106 60 decibels
82. 10 log
10I
83. 120 10 log2I
12
Difference 10 log
5
7
3.1610 10 10 log1.2610 10 12
12
10log3.16 107 log1.26 105 10 3.16 1.26 10
10 log
7 5
10log2.5079 102 10log250.79 24 dB
120 10log 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 logt 1, 0 ≤ t ≤ 12 (a) f t 90 logt 115
(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 log4 1 79.5
15 15 logt 1
(d) f 12 90 15 log12 1 73.3 (e)
75 90 15 logt 1 1 logt 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 54.40.964 t T 54.40.964 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
lnT 21
1T 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.
lnT 21 0.037t 4 T e0.037t4 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 lnx 2 2
90. f x
Properties of Logarithms
1 f x; false 2
f x ln x can’t be simplified further.
f x lnx ln x12
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 bxy v b Then logbuv logbb xy 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 log12 x
2
3
−3
log x ln x log 4 ln 4
6
−1
3
5
−3
−2
−3
log x ln x log12 ln12
6
−3
99. f x log11.8 x
98. f x log14 x
log x ln x log14 ln14
log x ln x log 11.8 ln 11.8
5
−1
−2
x ln x 101. f x ln , gx , hx ln x ln 2 2 ln 2 f x hx 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 ln2
2 ln 2 ln 2 0.6931 0.6931 1.3862
ln 5 1.6094 ln 6 ln2
3 ln 2 ln 3 0.6931 1.0986 1.7917
ln 8 ln 23 3 ln 2 30.6931 2.0793 ln 9 ln 32 2 ln 3 21.0986 2.1972
2 ln 5 ln 2 1.6094 0.6931 2.3025 ln 12 ln22 3 ln 22 ln 3 2 ln 2 ln 3 20.6931 1.0986 2.4848 ln 15 ln5 3 ln 5 ln 3 1.6094 1.0986 2.7080 ln 10 ln5
ln 16 ln 24 4 ln 2 40.6931 2.7724
2 ln 32 ln 2 2 ln 3 ln 2 21.0986 0.6931 2.8903 ln 20 ln5 22 ln 5 ln 22 ln 5 2 ln 2 1.6094 20.6931 2.9956
ln 18 ln32
103.
24xy2 24xx3 3x4 ,x0 16x3y 16yy2 2y3
105. 18x3y4318x3y43
107.
18x3y43 1 if x 0, y 0. 18x3y43
3x2 2x 1 0
104.
2x2
3
2x 3y
2
3
106. xyx1 y11
108.
3y3 27y3 2 3 2x 8x 6
xy x1 y1
xy 1x 1y
xy2 xy y xxy x y
4x2 5x 1 0
4x 1x 1 0
3x 1x 1 0 3x 1 0 ⇒ x
3y
4x 1 0 ⇒ x 14
1 3
x10 ⇒ x1
x 1 0 ⇒ x 1
The zeros are x 14, 1. 2 x 3x 1 4
109.
5 2x x1 3
110.
3x 1x 24
53 2xx 1
3x x 8 0
15 2x2 2x
2
x
1 ± 12 438 23
1 ± 97 6
0 2x2 2x 15 2 ± 22 4215 x 22 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. 42x7 64 425 7
3. extraneous
2. 23x1 32
x5
(a)
(b) x y (d) x
231 1
64
Yes, x 5 is a solution. x2
(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. 2e5x2 12
2e25 2
3e
25 3ee
e25
1 x 2 ln 6 5
(a)
75
2e5152ln 6 2 2e2ln 62
is not a solution.
2eln 6 2 6 12
x 2 ln 25
1 Yes, x 2 ln 6 is a solution. 5
3e2ln 25 2 3eln 25 325 75 Yes, x 2 ln 25 is a solution. (c)
x2 232 1 27 128
64
3. 3ex2 75
22 14
No, x 1 is not a solution. (b)
422 7 43
(a)
x 1
(a)
43
x 1.219
x
(b)
3e1.2192 3e3.219 75
ln 6 5 ln 2
2e5[ln 65 ln 2 2 2eln 6ln 2 2
Yes, x 1.219 is a solution.
2e2.5852 2 97.9995 195.999 No, x
ln 6 is not a solution. 5 ln 2 x 0.0416
(c)
50.0416 2
2e
2e1.792 26.00144 12
Yes, x 0.0416 is an approximate solution.
289
290
Chapter 3
Exponential and Logarithmic Functions
5. log43x 3 ⇒ 3x 43 ⇒ 3x 64
6. log2x 3 10
x 21.333
(a)
x 1021
(a)
321.333 64
log21021 3 log21024
Yes, 21.333 is an approximate solution.
Since 210 1024, x 1021 is a solution.
x 4
(b)
x 17
(b)
34 12 64
log217 3 log220
No, x 4 is not a solution.
Since 210 20, x 17 is not a solution.
x 64 3
(c)
364 3 64 Yes, x
64 3
x 102 3 97
(c)
log297 3 log2100 Since 210 100, 102 3 is not a solution.
is a solution.
7. ln2x 3 5.8
8. lnx 1 3.8 x
(a)
1 2 3
ln 5.8
x 1 e3.8
(a)
ln2 3 ln 5.8 3 lnln 5.8 5.8
ln1 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)
ln2 3 e5.8 3 lne5.8 5.8
ln45.701 1 ln44.701 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)
ln2163.650 3 ln 330.3 5.8
ln1 ln 3.8 1 lnln 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
2x 25
4x 43
x2
x5
x 5
x 3
x 5 13. ln x ln 2 0 ln x ln 2
ln x ln 5
x2
x5
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 e1
x e7
x 43
x 0.368
x 0.000912
x 64
20. log5 x 3 x 53 1
x 125 or 0.008
Section 3.4 21. f x gx
22. f x gx 27 9
2x 23
27x 2723 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 4x 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. 5t2 0.20 5t2
1 5
5t2 51 t 1 2 t2 2x3 32 x 3 log2 32 x35 x8
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.
ex2
5x 16
log3 3x log3 5
2xx 1 0
2 3
30. 25x 32
43x 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 ex ex
x5
x2 3 x 2
2x x 2 8
0 x 1x 2
31.
x41
Point of intersection: 9, 2
23, 9
0 x2 x 2
2
elnx4 e0
x9
x x2 2
28.
lnx 4 0
x 32
2 3
38.
43t 0.10 ln 43t ln 0.10
39.
3x1 33 x13
ln 0.10 ln 4
x4
t
ln 0.10 0.554 3 ln 4
ln 3000 ln 6 ln 3000 0.894 5 ln 6
3x1 27
3t ln 4 ln 0.10 3t
291
f x gx
24.
log3 x 2
x
x3
f x gx
23.
2 8 x
Exponential and Logarithmic Equations
292
Chapter 3
Exponential and Logarithmic Functions
23x 565
41.
82x 431
42.
ln 23x ln 565
ln 82x 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. 8103x 12
44. 510x6 7
12 8
103x
log 103x log
10 x6
32
ln 5x1 ln 7 7 5
x 1 ln 5 ln 7 x1
ln 7 ln 5
x1
7 x 6 log 5
ln 7 2.209 ln 5
6.146
0.059 836x 40
47. e3x 12
36x 5
48.
x
6 x ln 3 ln 5
e2x 50 ln e2x ln 50
3x ln 12
ln 36x ln 5
x
35x1 21 5x1 7
7 x 6 log 5
3 1 x log 3 2
6x
45.
7 5
log 10 x6 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
x6
ln 5 4.535 ln 3
49. 500ex 300 ex 35 x ln 35 x ln 35 ln 53 0.511
50. 1000e4x 75 3 e4x 40 3 ln e4x 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. 623x1 7 9
log2 23x1
462x 3.5
8 3
6 2x log4 3.5
3x 1 log2
55.
8462x 28
8 log2 3
x
6 2x
83 loglog832 or lnln832
1 log83 1 0.805 3 log 2
x3
ln 3.5 2.548 2 ln 4
ex 2ex 3 0 ex 5
(No solution)
57.
ln 3.5 ln 4
e2x 5ex 6 0
56.
ex 1ex 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 1ex 4 0
ex2 9ex 36 0
ex 10 ⇒ ex 1
Because the discriminant is 92 4136 63, there is no solution.
Not possible since ex > 0 for all x. ex 40 ⇒ ex 4 ⇒ x ln 4 1.386
59.
293
54. 8462x 13 41
623x1 16 23x1
Exponential and Logarithmic Equations
500 20 100 e x2 500 20100 e x2
60.
400 350 1 ex 400 3501 ex
25 100 e x2
8 1 ex 7
e x2 75 x ln 75 2
8 1 ex 7 1 ex 7
x 2 ln 75 8.635 ln
1 ln ex 7
x ln
1 7
x ln 71 x ln 7 x ln 7 1.946
61.
3000 2 2 e2x 3000 22 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 7e 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 ln1 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 ln1 0.10 12
67. gx 6e1x 25 Algebraically:
ln 30
15
6e1x 25
0.878 ln 30 26 t
6 −6
e1x
ln 30 0.409 3 ln16 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 4ex1 15
69. f x 3e3x2 962
20
0 4ex1 15 15 4ex1 3.75
ex1
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
3e3x2 962 e3x2
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 gx 8e2x3 11
70.
8e2x3 11
71. gt 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. ht e0.125t 8
e1.8x 7 0
Algebraically:
74. f x e2.724x 29 e2.724x 29
e0.125t 80
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
e2x3 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 e3 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
x2
e2
x e2 2 5.389
84. ln x 8 5 eln x8 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 e103 e103 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 e72 x e72 33.115 86. 2 6 ln x 10 6 ln x 8 ln x 43
x e23
eln x e43
0.513
x e43 0.264
296
Chapter 3
Exponential and Logarithmic Functions
87. 6 log30.5x 11
88. 5 log10x 2 11
log30.5x 11 6
log10x 2 11 5
3log30.5x 3116
10log10x2 10115
0.5x 3116
x 2 10115
x 23116 14.988
x 10115 2 160.489
89. ln x lnx 1 2
90. ln x lnx 1 1
x 1 2
lnxx 1 1
x e2 x1
xx 1 e1
ln
x
elnxx1 e1 x2 x e 0
x e2x 1
x
x e2x e2
1 ± 1 4e 2
x e2x e2 The only solution is x
x1 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 lnx 2 1
92. ln x lnx 3 1
lnxx 2 1
lnxx 3 1
xx 2
e1
elnxx3 e1
x2 2x e 0
xx 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 lnx 1 lnx 1
93.
lnx 5 ln x5
x1
x 1
x1 x1
x 5x 1 x 1 x2
6x 5 x 1
x2 5x 6 0
x 2x 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
lnx 1 lnx 2 ln x ln
x1
x 2 ln x x1 x x2 x 1 x2 2x 0 x2 3x 1
3 ± 32 411 x 21 3 ± 13 x 2 3.303 x (The negative apparent solution is extraneous.)
95. log22x 3 log2x 4 2x 3 x 4 x7
Section 3.4
Exponential and Logarithmic Equations
297
96. logx 6 log2x 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 27 1 are negative. There is no solution. 97. logx 4 log x logx 2
98. log2 x log2x 2 log2x 6
x4 logx 2 x
log2xx 2 log2x 6
log
xx 2 x 6 x2
x4 x2 x x 4 x2 2x 0
x2
x60
x 3x 2 0 x 3 or x 2
x4
1 ± 17 x 2
Quadratic Formula
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 x1 2
100. log3 x log3x 8 2
99. log4 x log4x 1
log3xx 8 2
3log3x
2 8x
32
x2 8x 9
4log4xx1 412 x 412 x1
x2 8x 9 0
x 9x 1 0
x 2x 1
x 9 or x 1
x 2x 2
The value x 1 is extraneous. The only solution is x 9.
x 2 x2 101. log 8x log1 x 2 log
8x 2 1 x 8x 102 1 x
8x 1001 x
2x 251 x 25 25 x 2x 25 25 x
2x 252 25 x
2
4x2 100x 625 625x 4x2 725x 625 0 x
725 ± 7252 44625 2529 ± 5 33 725 ± 515,625 24 8 8
x 0.866 (extraneous) or x 180.384 The only solution is x
2529 5 33 180.384. 8
298
Chapter 3
Exponential and Logarithmic Functions
102. log 4x log12 x 2 log
12 4x x 2
10log4x (12 x 102 4x 100 12 x 4x 10012 x 4x 1200 100 x 4x 1200 100 x x 300 25 x
x 3002 25 x
2
x2 600x 90,000 625x x2 1225x 90,000 0 x
1225 ± 12252 4190,000 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 lnx 2 10
y2 ln x
ln x 3
1 x 3
106. 10 4 lnx 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 ex2 3
2x 7 2x
500 1500ex2
−5
30 −1
lnx 2 2.5 elnx2 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 0.5e0.004x p 350
(a)
(b)
p 300
350 500 0.5e0.004x
300 500 0.5e0.004x
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 e0.002x
(a) When p $600:
(b) When p $400:
600 5000 1 0.12 1
4 4 e0.002x
400 5000 1
4 4 e0.002x
0.08 1
4 0.88 4 e0.002x
4 4 e0.002x
4 0.92 4 e0.002x
4 3.52 0.88e0.002x
4 3.68 0.92e0.002x
0.48 0.88e0.002x
0.32 0.92e0.002x
6 e0.002x 11
8 e0.002x 23
ln
6 ln e0.002x 11
ln
8 ln e0.002x 23
ln
6 0.002x 11
ln
8 0.002x 23
x
4 4 e0.002x
ln611 303 units 0.002
x
ln823 528 units 0.002
111. V 6.7e48.1t , t ≥ 0 (a)
(b) As t → , V → 6.7.
10
1.3 6.7e48.1t
(c)
Horizontal asymptote: V 6.7
0
1500 0
The yield will approach 6.7 million cubic feet per acre.
1.3 e48.1t 6.7 ln
67 13
t
48.1 t 48.1 29.3 years ln1367
300
Chapter 3
Exponential and Logarithmic Functions
112. N 68100.04x
113. y 7312 630.0 ln t, 5 ≤ t ≤ 12
When N 21:
7312 630.0 ln t 5800
21 6810
0.04x
630.0 ln t 1512
21 100.04x 68
ln t 2.4 t e2.4 11
21 log10 0.04x 68 x
t 11 corresponds to the year 2001.
log102168 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 e0.6114x69.71
50
1 e0.6114x69.71 2
1 e0.66607x64.51 2
e0.6114x69.71 1
e0.6667x64.51 1
0.6114x 69.71 ln 1
0.66607x 64.51 ln 1
0.6114x 69.71 0
0.66607x 64.51 0 x 64.51 inches
x 69.71 inches
116. P (a)
100 1 e0.66607x64.51
0.83 1 e0.2n (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 e0.2n e0.2n
0.83 1 e0.2n 0.83 0.60 0.83 1 0.60
ln e0.2n 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 201 72h (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
100 201 72h
(b) (b)
301
5 1 72h
200
4 72h 0
4 2h 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
ln47 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. logau v loga uloga v
119. logauv loga u loga v
False.
True by Property 1 in Section 3.3.
2.04 log1010 100 log10 10log10 100 2
121. logau 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 log100 10 log 100 log 10 1 124. A Pert
125. Yes.
(a) A 2P
ert
2
This doubles your money.
Pert
Time to Quadruple
2P Pe
4P Pert
(c) A Per2t Pertert ertPert
2 ert
4 ert
ln 2 rt
ln 4 rt
(b) A
Pertert
Time to Double
Pert
Pe2rt
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 25
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. gx
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. log34 5
log10 5 ln 5 5.595 log1034 ln34
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 aebx.
■
You should be able to use the Gaussian model y aexb c. 2
■
You should be able to use the logistic growth model a . y 1 berx
■
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 aebx
2. y a b ln x; y a b log x
4. bell; average value
5. sigmoidal
1. y 2ex4
3. normally distributed
3. y 6 logx 2
2. y 6ex4
This is an exponential growth model. Matches graph (c).
4. y 3ex2 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 lnx 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 e2x
ln 2 ln e0.105t ln 2 0.105t t
ln 2 6.60 years. 0.105
Amount after 10 years: A 750e0.10510 $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 750e0.08943810 $1834.37
ln 2 0.057762 5.7762%. 12
Amount after 10 years: A 10,000e0.05776210 $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
ln1505.00500 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
ln19,205600 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 Pe0.0210
Pe0.04510
P
10,000.00 P $6376.28 e0.04510 The time to double is given by t
15. 500,000 P 1 P
0.075 12
500,000 0.075 1220 1 12
500,000 $112,087.09 1.00625240
2000 $1637.46. e0.0210
The time to double is given by t
ln 2 15.40 years. 0.045
1220
ln 2 2 years. 0.3466
16.
AP 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% n1
(a)
n 12
(b)
1 0.11t 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 ln1 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 ln1
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 ln1 0.105
t
t
(c) n 365 365 ln1 0.105 365
3P Pert
19.
12 ln1 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 1.099r1. 21.
3P P1 rt
r
3 1 rt ln 3 ln1 rt ln 3 t ln1 r ln 3 t ln1 r
t
ln 3 (years) ln1 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 1.222r1.
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 Cek1599 2
25.
)
0.055 [[365t [[
A = 1 + 365
26.
1 C Cek1599 2
0.5 ek1599 ln 0.5 ln 0
ln 0.5 k1599
10 0
1 ek1599 2
ek1599
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 ek1599 2
ln
1 k1599 2 k
y 10e ln 0.51599 1000
ln12 1599
Given y 1.5 grams after 1000 years, we have
6.48 grams.
1.5 Ce ln121599 1000 C 2.31 grams.
27.
1 C Cek5715 2
28.
1 C Cek5715 2
0.5 ek5715
1 ek5715 2
ln 0.5 ln ek5715 ln 0.5 k5715 k
ln 0.5 5715
Given y 2 grams after 1000 years, we have 2 Ce ln 0.55715 1000 C 2.26 grams.
ln
1 ln ek5715 2
ln
1 k5715 2 k
ln12 5715
Given C 3 grams, after 1000 years we have y 3e ln125715 1000 y 2.66 grams.
29.
1 C Cek24,100 2 0.5 ek24,100 ln 0.5 ln ek24,100 ln 0.5 k24,100 k
ln 0.5 24,100
Given y 2.1 grams after 1000 years, we have 2.1 Ce ln 0.524,100 1000 C 2.16 grams.
Section 3.5
30.
1 C Cek24,100 2
y aebx
31.
1 ln ek24,100 2
ln
1 k24,100 2
1 1 aeb0 ⇒ a 2 2
10 eb3
1 5 eb4 2
ln 10 3b ln 10 b ⇒ b 0.7675 3
10 e4b ln 10 ln e4b
Thus, y e0.7675x .
ln12 k 24,100
307
y aebx
32.
1 aeb0 ⇒ 1 a
1 ek24,100 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 ln1224,100 1000
Thus, y 12e0.5756x.
C 0.41 grams. y aebx
33.
ln
y aebx
34.
5 aeb0 ⇒ 5 a
1 aeb0 ⇒ 1 a
1 5eb4
1 eb3 4
1 e4b 5
ln
15 4b
14 ln e
ln
4 3b
ln15 b ⇒ b 0.4024 4
3b
1
ln14 b 3
Thus, y 5e0.4024x.
⇒ b 0.4621
Thus, y e0.4621x .
35. P 2430e0.0029t (a) Since the exponent is negative, this is an exponential decay model. The population is decreasing.
(c) 2.3 million 2300 thousand 2300 2430e0.0029t
(b) For 2000, let t 0: P 2430 thousand people
2300 e0.0029t 2430
For 2003, let t 3: P 2408.95 thousand people ln
0.0029t 2300 2430 t
ln23002430 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
Canada
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
36. —CONTINUED— Canada:
(a) Bulgaria:
a 31.3
a 7.8
34.3 31.3eb10
7.1 7.8eb10 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 7.8e0.009430 5.88 million
y 31.3e0.0091530 41.2 million United States:
China:
ln
a 1268.9
a 282.3
1347.6 1268.9eb10
309.2 282.3eb10
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 1268.9e0.0060230 1520.06 million
y 282.3e0.009130 370.9 million
United Kingdom: a 59.5 61.2 59.5eb10 ln
61.2 10b ⇒ b 0.00282 59.5
For 2030, use t 30. y 59.5e0.0028230 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 10ek14
When t 3, y 10,000: 10,000 4080ek3 10,000 e3k 4080 ln
3k 10,000 4080 k
ln10,0004080 0.2988 3
When t 24: y 4080e0.298824 5,309,734 hits
ln
14k ⇒ k 0.1337 65 10
For 2010, t 20: y 10e0.133720 $144.98 million
Section 3.5 39.
N 100ekt
Exponential and Logarithmic Models
N 250ekt
40.
280 250ek10
300 100e5k 3 e5k
1.12 e10k
ln 3 ln e5k
k
ln 3 5k
500 250e ln 1.1210 t 2 e ln 1.1210 t
N 100e0.2197t
ln 2
200 100e0.2197t
41. R
ln 2 3.15 hours 0.2197
t
1 t8223 e 1012 R
(a)
ln
1 5715k 2
1012 814
k
t 1012 ln 14 8223 8
ln12 5715
The ancient charcoal has only 15% as much radioactive carbon. 0.15C Ce ln 0.55715 t
108 12,180 years old 12
14
ln 0.15
1 t8223 1 e 11 (b) 1012 13
t
1012 et8223 11 13
ln 2 61.16 hours ln 1.1210
1 C Ce5715k 2
1 814
t 8223 ln
ln 101.12t
y Cekt
42.
1 t8223 1 e 14 1012 8 et8223
ln 1.12 10
N 250e ln 1.1210 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 20
a 30,788
(b)
32,000
18,000 30,788ek2
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 30,788e0.268t
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 20
V 300t 1150 (c)
550 1150ek2
(b) ln
550 1150 2k ⇒ k 0.369
1200
V 1150e0.369t (d)
0
4 0
The exponential model depreciates faster in the first two years.
t
1
3
V 300t 1100
$850
$250
V 1150e0.369t
$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. St 1001 ekt 15 1001 ek1
(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) S5 1001 e0.16255 55.625 55,625 units
St 1001 e0.1625t 46. N 301 ekt (a)
N 19, t 20
N 25
(b)
25 301 e0.050t
19 301 e20k 30e20k 11 e20k
11 30
11 ln e20k ln 30 20k ln
11 30
k 0.050 So, N 301 e0.050.
5 e0.050t 30 ln
305 ln e
ln
305 0.050t
0.050t
t
ln530 36 days 0.050
Section 3.5 47. y 0.0266ex100 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. pt
1000 1 9e0.1656t
(a) p5
50. S
1000 203 animals 1 9e0.16565 500
(b)
(b) The average number of hours per week a student uses the tutor center is 5.4.
1000 1 9e0.1656t
(a)
500,000 1 0.6ekt 300,000
1 0.6e4k
5 3
0.6e4k
2 3
1 9e0.1656t 2 9e0.1656t 1 e0.1656t
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
ln19 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 0.6e0.02638
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 1012 wattm2. I0
(a) 10 log
1010 10 log 102 20 decibels 1012
(b) 10 log
105 10 log 107 70 decibels 1012
(c) 10 log
108 10 log 104 40 decibels 1012
(d) 10 log
1 10 log 1012 120 decibels 1012
312
Chapter 3
54. I 10 log
Exponential and Logarithmic Functions
I where I0 1012 wattm2 I0
(a) 1011 10 log
10 log
(b) 102 10 log
104 10 log 108 80 decibels 1012
(c) 104 10 log
55.
1011 10 log 101 10 decibels 1012
I I0
(d) 102 10 log
56.
I log 10 I0
102 10 log 1010 100 decibels 1012
10 log10 1010
10 10 10log II0 1010
102 10 log 1014 140 decibels 1012
I I0
I I0
I I01010
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
log2.3 105 4.64 59.
5.8 log H
log 11.3 106 4.95 60.
5.8 log H
3.2 log H
103.2 H
105.8 10log H
H 6.3 104 mole per liter
105.8 H
H 1.58 106 mole per liter 61.
2.9 log H
2.9 log H
H 102.9 for the apple juice 8.0 log H
8.0 log H
H 108 for the drinking water 102.9 108
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
10pH1 H
10pH1 H
10pH 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 0.07512
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 1.53851.0296s (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 gx 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
72. Answers will vary.
(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 12 5 22 12 32 10 (c) Midpoint:
(b) d 6 42 1 32
12 0, 2 2 5 21, 72
100 16 116 229 (c) Midpoint:
3 52 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 32 2 32 112 52 146 (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 22 172, 12
5 2 3 14 3 11
(b) d 10 72 4 02 9 16 25 5 (c) Midpoint: (d) m
7 2 10, 0 2 4 172, 2
4 40 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
14 0 14 1 (d) m 34 12 14
79. y 10 3x
232 73, 132 16 56, 121 13 16 12 1 23 73 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 2x 02 3
2
−6
x
−1
2
Parabola
2
(c) Midpoint:
(d) m
y
Slope: m 3
2
(b) d
12 2 34, 142 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 5x 6 2x 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 82 25
88. x 42 y 7 4
y
x 42 y 7 4
14
Circle
12
Center: 0, 8
x 42 y 3
10 8
Radius: 5
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 2x1 5
90. f x 2x1 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
93. Answers will vary.
Review Exercises for Chapter 3 1.
f x 6.1x
2.
f x 30x
3. f x 20.5x
f 3 303 361.784
f 2.4 6.12.4 76.699 4. f x 1278x5
5.
f 20.5 0.337
f x 70.2x f 11 70.211
f 1 127815 4.181
f x 145x
6.
f 0.8 1450.8 3.863
1456.529 7. f x 4x
8. f x 4x
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, gx 4x 3
11. f x 5x
Intercept: 0, 2
gx 5x1
Horizontal asymptote: y 1
Since gx f x 1, the graph of g can be obtained by shifting the graph of f one unit to the right.
Increasing on: ,
Because gx 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 , gx 8 3
1 13. f x 2
x
x
gx 12
x2
Because gx 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 gx 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 4x 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 2.65x1
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 5x2 4
18. f x 2.65x1
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 2x6 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
x2
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.
3x2 19
24.
3x2 32 x 2 2
13 x2 81 13 x2 34 13 x2 13 4
x 4
e5x7 e15
25.
e82x e3
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. e58 1.868
29. e1.7 0.183
30. e0.278 1.320
32. hx 2 ex2
31. hx ex2 x
2
1
0
1
2
x
2
1
0
1
2
hx
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 x2
34. st 4e2t, 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 3500e0.06510
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 2000e0.0530
n
1
2
4
12
365
Continuous
A
$8643.88
$8799.58
$8880.43
$8935.49
$8962.46
$8963.38
37. Ft) 1 et3 (a) F 12 0.154
(b) F2 0.487
(c) F5 0.811
3 38. Vt 14,000 4
t
(a)
39. (a) A 50,000e0.087535 $1,069,047.14
15,000
(b) The doubling time is
0
ln 2 7.9 years. 0.0875
10 0
(b) V2 14,00034 $7875 2
(c) According to the model, the car depreciates most rapidly at the beginning. Yes, this is realistic. 40. Q 10012
t14.4
(a) For t 0: Q 10012
014.4
(c)
100 grams
(b) For t 10: Q 10012
1014.4
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
2532 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 912 21
Review Exercises for Chapter 3 48. f x log4 x
47. gx log2 x g
1 8
f
log2 23 3
log2 18
1 4
log4 14
321
49. log4x 7 log4 14 1
x 7 14 x7
50. log83x 10 log8 5
51. lnx 9 ln 4
52. ln2x 1 ln11
x94
3x 10 5
2x 1 11
x 5
3x 15
2x 12 x6
x5 53. gx log7 x ⇒ x 7y Domain: 0,
3
gx
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
x1
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. gx log5 x ⇒ 5y x
y
−3
Vertical asymptote: x 0
3x ⇒ 3x 10
y
⇒ x 310 y
Domain: 0,
1 25
1 5
1
5
25
gx
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: x0
y
Domain: 0,
y
x-intercept: 3, 0
x
3
4
2
x 106
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 logx 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 logx 5 0 ⇒ logx 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 logx 3 1
y
Domain: 3, logx 3 1 0 logx 3 1 x3
x
4
5
6
7
8
f x
1
1.3
1.5
1.6
1.7
5 4 3 2 1 x −1
101
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 e12 12
62. ln e7 7
63. ln7 5 2.034
64. ln
65. f x ln x 3 6 5
x-intercept: ln x 3 0
4
ln x 3
4
lnx 3 0
2
x 3 e0
2
x 2
x4
1
e3, 0
y
Domain: 3,
3
x e3
66. f x lnx 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. hx lnx2 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
x1
−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 loga 40 176 h55 116 log55 40 176 53.4 inches
8
−2
x
Vertical asymptote: x 0
6
±4
Vertical asymptote: x 0
2.77
70. s 25
13 ln1012 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. log12 5
ln 200 2.132 ln 12
75. log 18 log2
log12 5
log 5 2.322 log12
32
76. log2
ln 0.28 1.159 ln 3
log3 0.28
1 log2 1 log2 12 0 log22 12
log 2 2 log 3
2 log2 22 log2 3 2
1.255
77. ln 20 ln22
log 0.28 1.159 log 3
74. log3 0.28
ln 5 2.322 ln12
3
log 3 log 2
3.585
5
78. ln 3e4 ln 3 ln e4
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
log33
1 log3 2
86. ln
log7 x12 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
lnx 3 ln x ln y
2 ln y 1 2 ln 4
lnx 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 lnx 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 x13
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 log8x 4 7 log8 y log8 8 3 3 log8 y7 x 4
323
y z2
90. 3 ln x 2 lnx 1 ln x3 lnx 12 ln x3x 12
92. 2 log x 5 logx 6 log x2 logx 65 log
x2 x 65
log
1 x2x 65
324
93.
Chapter 3
Exponential and Logarithmic Functions
1 ln2x 1 2 lnx 1 ln2x 1 lnx 12 2 ln
2x 1
x 12
94. 5 lnx 2 lnx 2 3 ln x lnx 25 lnx 2 ln x3 lnx 25 lnx 2 ln x3
lnx 25 ln x3x 2 ln
95. t 50 log
x 25 x3x 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 63
x3
x 3
101. log4 x 2
ln 6
103. ln x 4
61
x e4
x 4 16
6log6 x
x 16
x ln 6 1.792 ex 12
105.
x e3 0.0498
106.
ln ex ln12
e3x2 40 ln e3x2 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
14e3x2 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 1x 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. 45x 68
112. 212x 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 2ex 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
40.2x
118. 4e1.2x 9
16
Graph y1 25e0.3x 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. 25e0.3x 12
ex 4
x ln 2
Graph y1 −10
The graphs intersect at x 2.447.
or
116. 40.2x x 0
10
Graph y1 20.6x 3x.
e8.2
ln 95 1.833 ln 12
e2x 6ex 8 0
114.
e x 2e 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
e154 3
14.174
lnx 8 3
x 2 3
1 lnx 8 3 2
elnx3 e2
lnx 8 6
x e2 3
x 8 e6
ln
3x e154
124.
x e6 8 395.429
x 3e 22.167 2
326 125.
Chapter 3
Exponential and Logarithmic Functions
lnx 1 2
126. ln x ln 5 4
1 lnx 1 2 2
ln
lnx 1 4 elnx1
x 4 5 x e4 5
e4
x 5e4 272.991
x 1 e4 x e4 1 53.598 log8x 1 log8x 2 log8x 2
127.
log8x 1 log8 x1
128. log6x 2 log6 x log6x 5
x2
x 2
log6
x x 2 log x 5 6
x2 x5 x
x2 x2
x 2 x2 5x
x 1x 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.
x0 Since x 0 is not in the domain of log8x 1 or of log8x 2, it is an extraneous solution. The equation has no solution. 129. log1 x 1
130. logx 4 2
1 x 10
x 4 102
1
1 1 10 x
x 100 4
x 0.900
x 104
131. 2 lnx 3 3x 8
132. 6 logx 2 1 x 0
Graph y1 2 lnx 3 3x and y2 8.
Graph y1 6 logx 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 lnx 5 x 10 Graph y1 4 lnx 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. 37550 7550e0.0725t
134. x 2 logx 4 0
3 e0.0725t
Let y1 x 2 logx 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 logd 65
137. y 3e2x3
283 93 logd 65
Exponential decay model
218 93 logd
Matches graph (e).
logd
218 93
d 1021893 220.8 miles 139. y lnx 3
138. y 4e2x3 Exponential growth model
Logarithmic model
Matches graph (b).
Vertical asymptote: x 3 Graph includes 2, 0 Matches graph (f).
140. y 7 logx 3
141. y 2ex4 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 aeb0 ⇒ a 2 3
2eb4
1.5 e4b ln 1.5 4b
⇒ b 0.1014
Thus, y 2e0.1014x.
6 1 2e2x
y aebx 1 1 aeb0 ⇒ a 2 2 1 5 eb5 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 Ce250,000k 2
3499e0.0135t
4500 e0.0135t 3499
ln
1 ln e250,000k 2
0.0135t 4500 3499
ln
1 250,000k 2
t
ln45003499 18.6 years 0.0135
k
ln12 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 ln12250,000 5000 0.986C 98.6%C. After 5000 years, approximately 98.6% of the radioactive uranium II will remain.
147. (a) 20,000 10,000er5 2 e5r
2
40 ≤ x ≤ 100
1400 2000e3k
ln 2 5r
(a) Graph y1 0.0499ex71 128. 2
7 e3k 10
ln 2 r 5 r 0.138629
3k ln
13.8629% (b) A
149. y 0.0499ex71 128,
148. N0 2000 and N3 1400 so N 2000ekt and:
k
10,000e0.138629
$11,486.98
0.05
107
ln710 0.11889 3
40
100 0
(b) The average test score is 71.
The population one year ago: N4 2000e0.118894 1243 bats
150. N
157 1 5.4e0.12t
(a) When N 50: 50
(b) When N 75: 157 1 5.4e0.12t
75
1 5.4e0.12t
157 50
1 5.4e0.12t
5.4e0.12t
107 50
5.4e0.12t
e0.12t
107 270
e0.12t
0.12t ln t
107 270
ln107270 7.7 weeks 0.12
157 1 5.4e0.12t 157 75 82 75 82 405
0.12t ln t
82 405
ln82405 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 1016
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 1016
I 109.1 1,258,925,412
I 103.5 wattcm2
154. False. ln x ln y lnxy lnx 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 auv
6. f x 2 g x 2
e
ex 2
e
2 e2x e2x 2 e2x 4 4
4 4
au
av
f u f v (b) f 2x a2x ax2
(b)
6
y = ex
2x
e
x
2
ex 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 ex
9.
6
y4
y = ex
−6
y e x ex x
2
ey
3
ey
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 10
x x2 x3 x4 x5 . . . 1! 2! 3! 4! 5!
ex 1
y
4
ey
x ± x2 4 2
Quadratic Formula
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 1x ln . 2 2
11. Answer (c). y 61 ex 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
xay 1 ay 1 xay ay x 1 ayx 1 x 1 x1 x1 ln
x1 y loga x1
xx 11 ln a
f 1x
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 5001 0.07 1
1t
5001.07t
Simple interest formula: A Prt P 5000.07t 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
tk1
tk1
and y2 c2
1 c2 2
c1 1 c2 2
12
tk2
ln c1 ln c2 t t
1
k1 k1 ln12 2
1000
Simple Interest t 5 10 15 20 25 30
Time (in years)
B0 500
tk2 tk1
2
2000
200 500ak2
1 2
3000
14. B B0akt through 0, 500 and 2, 200
tk2
cc kt kt ln12
ln
4000
1
ln c1 ln c2 1k2 1k1 ln12
2 a2k 5 loga
25 2k
1 2 loga k 2 5 B 500a 12 loga 25 t 500 a log a 25 t2 500
25
t2
Problem Solving for Chapter 3 15. (a) y 252.6061.0310t
16. Let loga x m and logab x n. Then x am and x abn.
(b) y 400.88t 1464.6t 291,782 2
(c)
am
2,900,000
y2
amn
a b
amn1
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 logab x
ln x2 ln x2
17.
ln x2 2 ln x 0 ln xln 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 12x 12 y3 x 1 12x 12 13x 13 (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 12x 12 13x 13 14x 14
4
y = ln x
The pattern implies that ln x x 1 12x 12 13x 13 14x 14 . . . .
−3
9
y4 −4
20. y abx
y axb
ln y lnabx
ln y lnax 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 bx 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
y300 80.4 11 ln 300 17.7 ft3min
331
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
e65.411
9
0
Let x floor space in square feet and h 30 feet. 11,460 x30
65.4 11
x 382 cubic feet of air space per child. 23. (a)
(c) Total air space required: 38230 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 3.1141.341x 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 lnx
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: x35 8. 1
2. Solve for x: 3x1 81. 3. Graph f x 2x. 4. Graph gx 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: 72 49. 1
7. Solve for x: x 4 log2 64. 4 825. 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 log3x2 9. 14. Graph y lnx 2.
15. True or false:
ln x lnx y ln y
16. Solve for x: 5x 41 1 17. Solve for x: x x2 log5 25
18. Solve for x: log2 x log2x 3 2
19. Solve for x:
ex ex 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
