!

Unit 9– Exponential and Logarithmic Functions – Classwork In our study of Precalculus, we have examined polynomial expressions, rational expressions, and trigonometric expressions. What we have not examined are exponential expressions, expressions of the form y = a x . These types of expressions are very prevalent in the Precalculus theatre. An expression in the form of y = a x will graph an exponential. An exponential graph tends to “explode” based on the value of x, since the x is in the exponent.

!

y = 2x

y = 3x

y = .5 x

The graphs of various exponential curves are shown above. We know that the graph of y = 1x graphs a 1

! horizontal line. If a < 0, the graph of y = a!x will not exist at certain points for instance, ! what is ("2) 2 ? "2 x So it only makes sense to examine functions in form of y = a , if a >0, a " 1. When a >1, we get what is called a growth curve and the larger a is, the steeper the growth curve is. If 0 < a < 1, the we get a decay curve as shown in the 3rd graph above. No matter what, exponential curves in the form of y = a!x have certain features. ! What point do they have in common? (0,1) What is the domain? ("#,#) What is the range? (0,") Solving basic exponential equations can be accomplished by using the fact that if a x = a y , then x = y ! Examples) Solve for x.

2

x +1 x +1

!

! 2x"3

=8

1) 2 = 2 x =2

! 3

3

" 1 %1(x 3x +1 7 =$ ' ! # 49 &

5) 7 3x +1 = ( 7)(2(1(x) 3x + 1 = (2 + 2x x = (3

2x"3

2) 3

! 1` x +4 2

5

=9

= 125

1` x +4 2

2

=3

2x " 3 = 2 # x =

5 2

= 53 3) 5 1 x + 4 = 3 " x = #2 2 4 5x"1 = 3 32

9

2x"4

= (27)

2( 2x"4 )

x"1

!

3( x"1)

=3 6) 3 4 x " 8 = 3x " 3 x =5

5 3

2 2( 5x"1) = 2 7) 5 10x " 2 = 3

1 3 4 x"1 4) 3 = 3"1 4 x "1 = "1 x =0 34 x"1 =

!

30x " 6 = 5 # x =

11 30

8 5"2x = 1 8 5"2x = 8 0

8) 5 " 2x = 0 5 x= 2

! !

! 9. Exponentials and Logarithmic Functions

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!

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Solving exponential equations like the ones above are easy when each side of the equation have common bases. But problems like 3x "1 = 4 cause problems. With that problem created, we introduced the concept of logarithms. A logarithm is simply an inverse of an exponential. Students typically hear the word logarithm and go into a cold sweat because they do not understand them. So lets get it straight once and for all. The statement y = b x can be written in an alternate way: x = log b y . They mean the same thing. Whenever you are given a logarithmic statement, write it exponentially. You will know the answer!

!

Important: log b y = x is the same thing as saying b x = y

log 2 8

!

log 8 2

13) 8 x = 2

10) 3x = 81 x=4

log 25 3 5

!

1 2

14) 25 x = 5 1

2 3x = 2 2 " x =

!

log 3 81

!

9) 2 x = 8 x=3

log 9 0

15) 36 x = 6

5 2x = 5 3 " x =

!

12) 4 x = 32 2 2x = 2 5 " x =

!

6 2x = 6 2 # x =

18) 3x = "9 DNE

5 2

log 9 1

"1 2

"1

1 6

"1 4

16) 9 x = 1 x =0

2 " 1% $log 9 ' # 3&

!

5log 4 8

!

log 3 " 9

log 4 32

1 6

log 36

!

1 3

1

1 6

17) 9 x = 0 DNE

!

1 25 x 11) 5 = 5"2 x = "2 log 5

4x = 8

19) 2 2x = 2 3 3 15 x = ...5log 4 8 = 2 2

20)

9 x = 3(1 32x = 3(1 x=

2 (1 " 1% 1 ...$log 9 ' = 2 # 3& 4

! If the base is not specified, it is assumed to be 10. log 1 0 x and log x are the same things. Log base 10 are called ! common logs. !

Examples) Find the value of the following: log100

log1

x

22) 10 x = 1 x =0

21) 10 = 100 x =2

!

5log 10

"3log 3 10

!

1

24) 10 x = 10(3 2 " 1 % x = (3...$log ' =9 # 1000 &

log10 4

!

log10

26) 10 x = 10 3 27) 10 x = 10 4 x=4 1 x = .5...5log 10 = 2.5 x = .." 3log 3 10 = "1 3

25) 10 x = 10,5

9. Exponentials and Logarithmic Functions

!

2 " 1 % $log ' # 1000 &

1 log 10 x 23) 10 = 10"1 x = "1

!

-2-

!

!

5

28) 10 x = 10

5

x= 5

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!

In mathematics, we have a very special number and is called Euler’s number. Leonhard Euler (1707 - 1783) discovered this number and it is known as e. The value of e is 2.718281828…. e is a transcendental number which, like π and 2 , continues on forever without any pattern. (Note: the 1828 in e, although appearing twice consecutively near the start does not appear again for a very long while. It is completely coincidental that it appears twice). The number e is such an important number (if you would have to decide what the 5 most important numbers are, what!would they be? (0,1,",e,i) ), that it forms the basic of what are called natural logarithms of Napierian logs (after John Napier, 1550-1617, who first used them). Just as logarithms (log) use base 10, natural logs (ln) use base e. When you wish to find the value of a log, you write the expression exponentially. You do the same thing with a natural log except that your base is now e. ! For instance, to find ln 10, you call it x, and are now solving the equation e x = 10 . Since e is slightly below 3, we expect ln 10 to be between the values of 2 and 3. So, given the function y = e x , the domain is ("#,#) and the range is (0,") . Examples) Find the value of the following: 1 ln = x ! lne 4 = x e x 4 x 29) e = e 30) e = e"1

x=4

!

9lne

! ln e 31) e x = e.5 x = .5

x = "1

2

33) e x = e 2

!

1 1 ln 3 e3 34) e x = e"3

ln1

!

35)

32) e x = 1 x =0

!

ln("2e)

e ln 5 36) ln5 = x

DNE 1 1 e x = 5 " e ln 5 = 5 x = "3... ln 3 = "1 3 e There are four basic rules for operation with logarithms that you must know. They are as follows: ! ! ! "$ a % log b x b ! = log a ' log b 1. log (a " b) = log a+ log b 2. log 3. log a = blog a 4. log a x = # b& log b a

x = 2...9lne 2 = 18

These rules work with logs to any base or the ln function. The last rule is called the change of base formula. Examples) Find the value of the following expressions: !

3 32 log 2 + log 2 38) 2 3 log 2 16 = 4

log25 + log 4 37) log100 = 2

!

!

39)

log 2 40 " log 2 5 log 2 8 = 3

40)

log 5 100 " log 5 4 log 5 25 = 2

4 " log 1 12 3 3 3 log10 ! ! 41) 44) ! 4 35log10 = 35 log 1 =2 2 3 36 For problems 45 – 49 – use the change of base formula (#4) with base 10 and calculate to 3 decimal places. log 4 20 log 2 36 log 5 100 log 3 0.5 ! ! 45) log20 46. log 36 47) log100 48) log.5 = 2.161 ! = 5.170 = 2.861 ! = "0.631 log 4 log2 log5 log 3 5

( 2) 42) 5 5log ( 2 ) = 2 log 2

35

log50 + log 4 + log5 ! 43) log1000 = 3

9. Exponentials and Logarithmic Functions

!

!

-3-

!

log 1

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!

Solving logarithmic equations Solving logarithmic equations (statement with logs in them) is simply a matter of writing them exponentially and solving via conventional methods. The rules you just studied can be quite helpful as well. log 5 (2x + 5) = 2

log 3 ( 4 x " 7) = 4

49) 25 = 2x + 5 x = 10

50) 81 = 4 x " 7 x = 22 log 4 ( x 2 " x + 2) =

log x (2x + 8) = 2

!

51)

!

x 2 " 2x " 8 = 0

( x " 4 )( x = 2) = 0 x = 4, x = "2/

53)

log 4 ( x "1) = 2

log 6 ( 35x + 6) " log 6 x = 2 !

4 x " 4 = 100 x = 26 Solving Exponential Equations

!

35x + 6 =2 x 54) 35x + 6 36 = x x =6 log 6

Exponential equations are in the form of a f ( x) = b or a f ( x) = b g ( x ) . To solve such equations, the technique is to ! take the log (any base but 10 is best) of both sides. And use the fact that log a b = blog a . Solve for x algebraically and then 3 decimal places and verify using the calculator, either numerically or graphically. ! x 3.4 x +4 = 29.6 3 = 12 55) x log 3 = log12 log12 x= = 2.262 log 3

!

2 52) x " x + 2 = 2 x ( x "1) = 0

x = 0, x = 1

log( x "1) + log 4 = 2

!

1 2

10000 = 4000(1.06)

56) ( x + 4 ) log 3.4 = log29.6

x=

3x = 12 x"2

x

5 57) = 1.06 x 2 log2.5 x= = 15.725 log1.06

log29.6 " 4 = "1.232 log 3.4

!

x log 3 = ( x " 2) log12 58) x log 3 = x log12 " 2log12

x=

2log12 = 3.585 log12 " log 3

2e 2x"5 = 24

!

e x = 10 59) x = ln10 = 2.303

!

9. Exponentials and Logarithmic Functions

!

e 2x"5 = 12

60) 2x " 5 = ln12 5 + ln12 x= = 3.742 2 -4-

!

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Exponential Growth When things grow exponentially, they “explode”, meaning that as time grows, the number of them increase slowly at first and then faster and faster. When we have exponential decay, the number of them decreases quickly at first and then slower and slower. Here are the graphs of exponential growth and decay:

Let’s give some examples of real-life applications of exponential growth and decay. Growth: Money, Populations, Antiques, speeds of computers Decay: Diseases, half-life of elements Exponential growth/decay can be modeled by the equations below. You are responsible for these equations.

Exponential Growth P = po (1+ r)

!

Exponential Decay

t /n

P = po (1" r)

t /n

where: po = the initial population (the amount at time zero) r = the rate of change (a decimal ! greater than zero) n = number of time periods it takes for the rate to occur t = time - it will be the x variable when using the calculator ! P = the new population - it will be your Y when using the calculator 1. Assume that the number of bacteria in a bacterial culture doubles every hour and that there are 1,000 present initially. Variables: po = 1000

r = 100%

n=1

a. Write the equation that describes this growth pattern. P = 1000(1+ 1)

t

b.! Draw a graph that shows the number of bacteria present during the first 12 hours. ! c. How many bacteria are present after 8 hours and 15 minutes?

304,437

d. Determine mathematically when the number of bacteria will be 1 million. Verify graphically. t 1000000 = 1000(2) 1000 = 2 t 3 t= = 9.966 years log2 9. Exponentials and Logarithmic Functions

!

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2. For the same problem, suppose that the number of bacteria triples every 2 hours. How does that change the problem? Variables: po = 1000

r = 200%

n=2

a. Write the equation that describes this growth pattern. P = 1000(1+ 2)

t

2

b.! Draw a graph that shows the number of bacteria present during the first 12 hours. ! c. How many bacteria are present after 8 hours and 15 minutes?

93,923

d. Determine mathematically when the number of bacteria will be 1 million. Verify graphically. 1000000 = 1000(1+ 2)

t

2

t

1000 = 3 2 3 t =2 = 12.575 years log 3

3. The population of a town is 100,000 and increasing at the rate of 2.15% a year. Variables: po = 100000

r! = .0215

n=1

a. Write the equation that describes this growth pattern. P = 100000(1+ .0215)

t

b.! Draw a graph that shows the population of this town during the first 50 years. ! c. How many people are present after 10 years?

123,704

d. Determine mathematically when the population will have doubled. Verify graphically. t 200000 = 100000(1+ .0215) 2 = 1.0215 t log2 t= = 32.585 years log1.0215

4. The population of Allentown, Pa. was 5,000 in the year 1900 and has steadily grown by 12.68% every 4 years since. ! Variables: po = 5000

r = .1268

n=4

a. Write the equation that describes this growth pattern. P = 5000(1+ .1268)

t

4

b.! Draw a graph that shows the population of this town during the 20th century. c. How many people live in Allentown in the ! year 2008? d. Determine when the population will reach 150,000. 9. Exponentials and Logarithmic Functions

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5. The population of a small, western town in the year 1860 was 7,500 and decreasing at the rate of 3.85% every two years. Variables: po = 7500

r = -.0385

n=2

a. Write the equation that describes this decay pattern. P = 7500(1" .0385)

t

2

b.! Draw a graph that shows the population from 1860 to the present. c. How many people lived in the town in the!year 1935. d. How many people live there today?

1721

410

e. In what year was the population cut in half? Do mathematically and verify graphically. t t 1 3750 = 7500(1" .0385) 2 = .9615 2 t = 35.3 year 1895 2 6.

The half life of carbon-14 is 5,730 years. Assume that a typical male has 35 pounds of carbon-14 in his body at death. ! t a. Write the equation that describes this decay pattern. P = 35(.5) 5730 b. Draw a graph that shows the amount of carbon-14 left after 10,000 years.

! 1,000 years. c. How much carbon-14 will be remaining after

31.012 lbs

d. How many years will it take to have only 1 pound of carbon-14 left?

29,391

e. If Jesus Christ's body were found today, how much carbon-14 would be remaining? 27.452 lbs. (Answer for the year 2008). 7. Not all half-lives are that long. Some elements have relatively small half-lives. In the following chart are some basic elements and their half-lives. If you started with 1,000 grams of the element, determine would remain in two minutes. Element Neon-23 Oxygen-10 Beryllium-11 Nitrogen-16 Carbon-15

Half-life 37.24 seconds 26.91 seconds 13.81 seconds 7.13 seconds 2.5 seconds

Remaining after 2 minutes 107.147 gm 45.458 gm 2.422 gm .009 gm -12 3.553(10) gm

To see these half-lives in action, check out the applet at http://www.colorado.edu/physics/2000/isotopes/radioactive_decay3.html !

9. Exponentials and Logarithmic Functions

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One of the basic uses of exponential growth is that of money and investments. Money gains interest while sitting in a bank and people taking out loans have to pay interest on those loans because if the money were still in the bank, it would gaining interest. There are two types of interest, simple and compound interest. Simple interest is exactly that, you put a sum of money in the bank (principal) and gain the same amount of interest every year. Almost no bank gives simple interest. One example of interest is a parent giving a loan to a child. If a parent loaned his child $5,000 to buy a car and charged him 4% interest over 10 years, the amount the child will pay back is according to the formula: P = P0 + P0 rt where P0 is the principal, r is the rate of interest and t is the time in years. What is the payback and how much is the yearly interest?

Compound ! interest is the way all banks give interest. Compound interest means interest upon interest. For example, assume we put $5,000 in the bank at 4% interest. Let’s determine the amount of interest for each year. Principal ( P0 ) 5,000 5,200 5,408 ! 5,624.32 5,649.29 6083.26 6,326.59 6,579.65 6842.84 7,116.55

Interest in year: ( P0 r ) 1:200 2:208 3:216.32 ! 4:224.97 5:233.97 6:243.33 7:253.06 8:263.19 9:273.71 10:284.66

New Principal 5,200.00 5,408.00 5,624.32 5,849.29 6,083.26 6,326.59 6,579.65 6,842.84 7,116.55 7,401.21

Note how the interest increases each year. Note that the final principal is much more than the $7,000 we had with simple interest. Rather then completing this chart for longer periods of times, we use the following formula for compound interest. Since it is an example of exponential growth, it has a similarity to the formula we used in the last section.

" Formula for compound interest: P = P0 $1+ # where:

!

r% ' n&

nt

P0 = initial principal (principal at time = zero) r = Annual Percentage Rate (APR) – this is what banks publish – your rate of interest yearly ! in years t = time n = number of compounding periods – banks give interest in varying number of periods bi-annually – interest every two years (n = .5) – quite rare annually – interest once a year (n = 1) – somewhat rare semi-annually - interest twice a year (n = 2) – somewhat rare quarterly – interest every 3 months (n = 4) – somewhat common monthly - interest every month (n = 12) – some banks do it weekly – interest every week (n = 52) – quite rare daily – interest every day (n = 365) – most banks do it

9. Exponentials and Logarithmic Functions

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8. If I invested $5,000 in a bank giving 4% interest for 10 years, how much would I have according to the following compounding methods? Set your calculators to 2-decimal places. Bi-annual 7,346.64

Annual 7,401.22

Semi-annual 7,429.74

Quarterly 7,444.32

Monthly 7,454.16

Weekly 7,457.98

Daily 7,458.96

Every hour 7,459.12

Every minute 7,459.12

Every second 7,459.12

Continuous Compounding Note that as n gets larger, there is essentially no difference in the amount of interest you will gain. Some banks compound continuously. The formula for continuous compounding is

P = P0e rt

!

where P0 is the initial principal, r is the annual rate of interest and t is time in years. Here is our first real-life use of the number e which you recall is 2.718281828… You have an e button on your calculator - in fact you ! have two of them. The first is above the division sign and the second is above the ln key which allows for taking e to a power. Suppose you invested $1 in a bank at 5% interest as a child and simply left it there to gain interest. If it compounded interest according to the continuous compounding formula, how much would be in the bank after: 1 year 1.05

10 years 1.65

50 years 12.18

100 years 148.41

200 years 22,026,47

500 years $72 billion

9. You plan to invest $10,000 in a CD (certificate of deposit) for 5 years. There are four CD’s available to you with the following rates and compounding. Which one give you the best deal and why? a. 5.15% - annually 12,854.24

b. 5.1% - quarterly 12,883.83

c. 5.05% - monthly 12,865.58

d. 5.03% - daily 12,859.31

10. At the birth of their child, Mr. and Mrs. DeGenerous wish to give him a new car (worth $20,000) upon his graduation from high school when he is 18 years old. How much should they invest at 5.25% APR compounded daily in order to have the money available to this boy upon graduation?

" .0525 % 365(18) P$1+ = 20000 ' # 365 &

P=

20000 = $7,774.12 " .0525 % 365(18) $1+ ' # 365 &

11. George Dubya wants to double his money in 10 years. Find the rate of interest he would need if his money compounds continuously. !

Pe10r = 2P 9. Exponentials and Logarithmic Functions

!

e10r = 2

10r = ln2 " r = -9-

ln2 = 6.93% 10 www.mastermathmentor.com - Stu Schwartz

Exponential Curve Fitting If we have two points, we know that we have one line that goes through them. We can find the equation of this line using the equations y = mx + b or y " y1 = m( x " x1 ) . We can also find the equation of the exponential curve y = Ce ax that fits these two points. Example 1) Find the equation of both the line and exponential curve going through the point (4, 6) with ! y-intercept = 3. For both curves, if x = 8, find the corresponding y-value.

!

Line work: (0,3), ( 4,6)

Exponential work: (0,3), ( 4,6)

3 m= 4 ! 3 y= x+3 4 Corresponding y-value (8, 9)

y = Ce ax 3 = Ce 0 " ! C = 3 so y = 3e ax 6 = 3e 4 a " 2 = e 4 a

4a = ln2 " a =

ln 2 4

Corresponding y-value (8, 12)

When you calculate the value of a on ! your calculator, be sure to store it. The calculator’s accuracy Setting doesn’t matter as long as you store the value. The screens below show the process.

Example 2) Find the equation of both the line and exponential curve going through the point (3, 5) with y-intercept = 12. For both curves, if x = 12, find the corresponding y-value. Line work: ( 3,5), (0,12)

Exponential work: ( 3,5), (0,12)

7 m=" 3 ! 7 y = " x + 12 3

y = Ce ax 12 = Ce 0 " C!= 12 so y = 12e ax 5 = 12e 3a "

5 = e 3a 12

3a = ln 125 " a =

ln 125 = #.2918 3

Corresponding y-value (12, -16) Corresponding y-value (12, 0.3617) ! ! 3) Mountain Training: It is determined that a normal person cannot survive barometric pressure less than 15. The barometric pressure at sea level is 29.5. At Mile High Stadium in Denver, the pressure is 24.5. Assuming an exponential model, could someone survive Mt. Everest (29,028 feet) without training? What is the highest mountain someone could survive?

(0,29.5), (5280,24.5) y = Ce at

29.5 = Ce 0 " C = 29.5 so y = 29.5e at

24.5 = 29.5e 5280a "

24.5 = e 5280a 29.5

9. Exponentials and Logarithmic Functions

!

5280a = ln

ln 24.5 24.5 " a = 29.5 29.5 5280 - 10 -

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4) Marijuana: Some doctors use a system to determine the amount of marijuana in the bloodstream. After smoking a joint of marijuana, the scale reads 100. Six hours later, the scale will be at 78. Assuming an exponential model, what will be the scale reading a) 12 hours later, b) 24 hours later, c) 1 week later. If someone’s MJ scale read 54 at 12 midnight, estimate the time when they last smoked marijuana.

(0,100), (6,78) y = Ce at

100 = Ce 0 " C = 100 so y = 100e at ln.78 78 = 100e 6a " .78 = e 6a 6a = ln(.78) " a = = #.0414 6 y (12) = 60.84 y (24 ) = 37.02 y (168) = 0.10 Using the graph, it took 14.88 hours for the scale to reach 54 so the joint was smoked about 15 hours earlier or just after 9 AM. 5. Fruit Fly Growth: Suppose that an experimental population of fruit flies increases according to the law of continuous exponential growth: If we start with 100 fruit flies, and have 300 flies after the 2nd day, how many will we have after 10 days? Calculate mathematically how long it will take for there take to be a billion flies? Why won’t the world get inundated with flies?

!

1000000000 = 100e at

(0,100), (2,300)

10000000 = e at

y = Ce at

100 = Ce 0 " C = 100 so y = 100e at ln10000000 at = ln10000000 " t = = 25.1 days a ln 3 2a 2a 300 = 100e " 3 = e 2a = ln 3 " a = = .5493 Exponential growth doesn't continue forever. 2 y (10) = 24,300 flies The more flies we have, the more will die off of hunger and crowding. !

6. Local Population: North Wales had 4,500 people living there in 1980. The population grows exponentially ! Remember the exponential growth formula: and in 1995, there were 5,100 people living there. a) How many people are living there in 2008? b) In what year will there be 7,500 people living there?

(0,4500), (15,5100) y = Ce at

4500 = Ce 0 " C = 4500 so y = 4500e at

17 15a =e 15 y (28) = 5,684 people

5100 = 4500e15a "

15a = ln

ln 17 17 " a = 15 = .0083 15 15

As seen on the graph, it will take 61 years for the population to reach 7,500. It will happen in the year 2041.

! 9. Exponentials and Logarithmic Functions

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Using the Calculator for Financial Operations (Optional) The TI-83 contains built-in commands that allow for calculation of values associated with compound interest, annuities, amortized loans and payout annuities. The Finance menu is opened by pressing [2nd] [x-1] on the TI-83, or by pressing the [APPS] key and selecting 'Finance' on the TI-83 plus or TI-84. The first menu option under Calc is the TVM-Solver. This is the primary mechanism by which we will perform financial calculations. The variables listed in the TVM solver are called 'fields'. Each variable represents a quantity associated with a common financial concept or formula. You will fill in all of these fields except one of them which is the one you are trying to find. You place your cursor on that field, then press the SOLVE button which is ALPHA ENTER. • N: This represents the number of compounding periods in the term of the investment, annuity or loan. This will always be a positive integer. • I%: This represents the 'nominal rate' for an investment, annuity or loan. This will always be a positive integer. Note: We write the percent form here, not the fraction or decimal form of a percent. • PV: This represents the 'present value' of an investment, loan or annuity. This number can be positive or negative. if the number is positive, then it indicated money we collected as in a loan. If the number is negative, then it represents money we paid out, as in an investment or loan where we are the lender. • PMT: This represents the payment made to build an annuity or pay off a loan. The value will always be negative in these situations. If we have a 'payout' annuity, then the value will be positive. In either case, the value represents the payment per compounding period. • FV: This represents the 'future value' of an investment, annuity or loan after N compounding periods have passed. This value will be positive or negative depending on the signs of PV and PMT. • P/Y: This value represents the number of payments per year for annuities and loans. • C/Y: This represents the number of compounding periods per year. These must both be positive integers greater than 1. • PMT: END BEGIN. This field allows one to set the TVM Solver for 'ordinary' annuities, (END), or annuities 'due' (BEGIN). For instance, suppose we wanted to invest $5,000 at 5.75% interest for 10 years compounded daily. 365(10) " 0575 % P = 5000 1+ . = 8885.25 . To setup the TVM Solver; Our formula from the last section would be: $ ' # 365 &

N = 3650 as there are 365 compounding periods in a year and 10 years. The interest rate is 5.75%. Note that it is not entered as .0575. The Present Value (PV) is -5000. It!is negative as we are paying it out to the bank. Payment (PMT) is 0 as we pay nothing beyond the initial $5.000. The Future Value is what we are trying to find. You will place your cursor on that field. Make both P/Y and C/Y = to 365. Payments are made at the end of the payment period (every day). 9. Exponentials and Logarithmic Functions

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When your press SOLVE you get that the Future value of the loan is $8,885.25. Sam Saver gets paid every 2 weeks and puts $100 from his paycheck into an annuity paying 6% interest compounded daily. If he continues this practice for 5 years, how much money will he have in the annuity? N = 5(26) = 130 (26 pay periods per year) Interest rate is 6% The Previous value is 0 – no money in the account. The payment is – 100 as we are paying it to the bank. The Future Value is what we are trying to find. You will place your cursor on that field. The payments per year is 26. The compounding periods per year is 365. When your press SOLVE you get that the Future value of the loan is $15,142.87. He has paid in $13,000 and thus has made $2,142.87 in interest or about 16%. Suppose you purchased a house for $400,000 at 5.25% and wanted a 30 year mortgage. What would your monthly payment be? N = 360 as you are making monthly payments for 30 years. Interest rate is 5.25%. The Present Value is positive as it is a loan. We want the payment so we put our cursor on this field. The Future Value is zero as we want to owe nothing eventually. Both P/Y and C/Y are 12 as we are making 12 payments a year. PMT is BEGIN as mortgage payments are made at the beginning of every month. So we find that our monthly payment is $2,199.19. Note that if we make 360 payments over 30 years, we pay back $791,708.40. Since the house cost $400,000, we are paying $391.708.40 in interest payments. In this case the interest you pay over 30 years is just about the cost of the house. So let’s suppose that you cannot afford to pay $2,199.19 a month. You examine your financial situation and decide that the most you can pay is $1,575 a month. How expensive of a house can you afford? Change the Payment to -1,575. Place the cursor over PV (it doesn’t matter what is in there as it will be recalculated) and press SOLVE. We can buy a house for $286,468.57. Now suppose you find a house for $300,000 that you like. You still cannot pay more than $1,575 a month. What interest rate would you need in order to afford the house? Change the Present Value to 300000 and put your cursor over the interest rate. It makes no difference what is in there as it will be recalculated. Press SOLVE. We need to find an interest rate of 4.84%

9. Exponentials and Logarithmic Functions

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www.mastermathmentor.com - Stu Schwartz

Unit 9 - Exponential and Logarithmic Functions - Homework 1. For each curve below, identify it by the proper equation number. Check out certain known points (like x = 0) to do a trial and error process. a. y = 2 x

b. y = 2"x

c. y = 2(2 x )

d. y = "3x

e. y = "3(2 x )

f. y = .5( 4 x )

g. y = 4 x " 2

h. y = .5 x " 3

i. y = "2(.5 x ) + 1

!

j. y = 3x +1

!

k. y = 2 x"1 + 2

!

!

m. y = 2

n. y = 1.1x

!

!

! !

!

!

! !

!

!

!

x

2

l. y = 0.5 x"2 o. y = 4 x + 4 "x

1. d

2. m

3. i

4. k

5. a

6. l

7. b

8. o

9. g

10. h

11. c

12. j

13. n

14. f

15. e

9. Exponentials and Logarithmic Functions

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www.mastermathmentor.com - Stu Schwartz

Solve for x 2 x"3 = 16

!

!

16. 2 x"3 = 2 4 x "3= 4 # x = 7

17. 32x"3 = 34

1 2 5"2x = 2"1 19. 2 5 " 2x = "1 # x = 3

1 100 5x +6 = 10"2 20. 10

27

22. 3

16

=3

9

# 1 & x"3 =% ( $ 27 '

2( 2x"4 )

#7 9

!

"3( x"3)

=3 25. 3 4 x " 8 = "3x + 9 17 7x = 17 ) x = 7

!

log 4 256

!

3( x"3)

" 1 % x(7 " 1 % x(11 $ ' =$ ' # 32 & #8&

log 2 8

!

31. 5 x = 125 x=3

32. 32x = 33 3 x= 2

!

34. 25 x = 5 1 x= 2 9. Exponentials and Logarithmic Functions

!

24. 5

1 #1 "x= 2 8

2( 6"2x )

1 2

=5 1 23 12 " 4 x = # x = 2 8

" 1 % 2(2x = $ ' #4&

! 27.

(2( 2(2x )

( 2) 3

3x +6

1 ( 3x +6) 3

2 =2 (4 + 4 x = x + 2 3x = 6 ) x = 2

!

log 9 27 9 x = 27

=2

1 2

25 6"2x = 5

=2 23. 2 4 x "12 = 3x " 9 # x = 3

log 5 125

4 x +1

4x +1=

x"3

29. 2 x = 8 x=3

!

!

=8

4 ( x"3)

"8 5

(5x + 35 = (3x + 33 2x = 2 ) x = 1

!

log 25 5

21. 2

26. 2(5( x(7) = 2(3( x(11)

28. 4 x = 256 x=4

!

x"3

2 4 x +1 = 2

!

5x + 6 = "2 # x =

!

2

9x + 9 = 2 " x =

2x"4

!

18. 5 3x"3 = 5 0 3x " 3 = 0 # x = 1

3 2

10 5x +6 =

=9

3( 3x +3)

!

2x " 3 = 4 # x =

2 5"2x =

3x +3

5 3x"3 = 1

32x"3 = 81

!

log 8 2

30. 8 x = 2 1 x= 3

log 7 1

33. 7 x = 1 x =0

!

log 8 16 8 x = 16

log 3 27

35. 2 3x = 2 4 4 x= 3

36. 3.5x = 33 x =6 - 15 -

www.mastermathmentor.com - Stu Schwartz

!

1 125 5

log 1

37. 5"1x = 5"3 x=3

!

38.

lne 7

!

40. e x = e 7 x=7

10 log 29 39. log29 = x

"5log12 12 5(1) = "5

e ln

10 x = 29

e

41. e x = e ln

!

e

x= e

3log10 2lne 42. 3 2

Solve each equation in terms of x.

!

!

log 3 (2x " 2) = 2

log 7 ( 7x " 6)!= 2

43. 2x " 2 = 9 11 x= 2

!

44) 7x " 6 = 49 55 x= 7

log 5 ( x + 3) " log 5 x = 2

ln 3x + ln 3 = 3 ln9x = 3

!

45) e 3 = 9x x=

e3 9

x+3 =2 x 46) x+3 25 = x 1 x= 8 log 5

!

!

log 2 ( x "1) + log 2 ( x + 3) = 5 47) x 2 + 2x " 3 = 32

( x + 7)( x " 5) = 0 x = "/ 7/ ,5

Use the change of base formula using base 10 to calculate the following to 3 decimal places. ! log 6 50 49) log50 = 2.183 log6

!

4 x + 22 =1 2x + 1 4 x + 22 = 10 48) 2x + 1 20x + 10 = 4 x + 22 3 x= 4 log

log 2 ( x "1)( x + 3) = 5

!

log( 4 x + 22) " log(2x + 1) = 1

log 2 25 50) log25 = 4.644 log2

log12 24 51) log24 = 1.279 log12

! 9. Exponentials and Logarithmic Functions

! - 16 -

log 4 3 52) log 3 = 0.792 log 4

! www.mastermathmentor.com - Stu Schwartz

Solve for x algebraically and then 3 decimal places. Verify using the calculator, either graphically or numerically.

6 x"2 = 15

5 x = 10 53) x log5 = 1 1 x= = 1.431 log5

!

54) ( x " 2) log6 = log15

x=

5 2x"5 = 20 55) 2x log5 " 5log5 = log20

x=

!

x log6 = ( x + 3) log 4 56) x log6 = x log 4 + 3log 4

5log5 + log20 = 3.431 2log5

x=

50 2x +5 = 2 3x"1

!

(2x + 5) log50 = (3x "1) log2 57) 2x log50 + 5log50 = 3x log2 " log2

x=

6 x = 4 x +3

!

(2x " 5) log5 = log20

log15 + 2 = 3.511 log6

5log50 + log2 = "3.526 3log2 " 2log50

3log 4 = 10.257 log6 " log 4

e 2x +1 = 100 58) 2x + 1 = ln100 ln100 "1 x= = 1.803 2

Exponential Growth Problems

!

! 1. Assume that the number of bacteria in a bacterial culture triples every hour and that there are 20 present initially.

Variables: po = 20

r=2

n=1

a. Write the equation that describes this growth pattern. P = 20(1+ 2)

t

b.! Draw a graph that shows the number of bacteria present during the first 12 hours. ! c. How many bacteria are present after 6 hours and 45 minutes. 33,235 d. Determine mathematically when the number of bacteria will be 1 million. Verify graphically. 1000000 = 20(1+ 2)

t

50000 = 3t log50000 t= = 9.849 hours log 3

9. Exponentials and Logarithmic Functions

!

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www.mastermathmentor.com - Stu Schwartz

2. For the same problem, suppose that the number of bacteria doubles every 3 hours. How does that change the problem? Variables: po = 20

r=1

n=3

a. Write the equation that describes this growth pattern. P = 20(1+ 1)

t

3

b.! Draw a graph that shows the number of bacteria present during the first 12 hrs. c. How many bacteria are present after 6 hours!and 45 minutes?

95

d. Determine mathematically when the number of bacteria will be 1 million. 1000000 = 20(1+ 1)

t

3

t

50000 = 2 3 3log50000 t= = 46.829 hours log2

3. The population of a city is 1,000,000 and increasing at the rate of 1.97% a year. Variables: po = 1,000,000! r = .0197

n=1

a. Write the equation that describes this growth pattern. P = 1000000(1.0197)

t

b.! Draw a graph that shows the population of this town during the first 60 years. ! c. How many people are present after 10 years?

1,215,414

d. Determine mathematically when the population will have doubled. t 20000000 = 1000000(1+ .0197)

2 = (1.0197) t=

t

.

log2 = 35.531 years log1.0197

4. The population of New York, NY was 7.9 million in the year 1970 and has averaged 3.1 % growth every 10 years since. ! Variables: po = 7.9

r = .031

n = 10

a. Write the equation that describes this growth pattern. p = 7.9(1+ .031)

t 10

b.! Draw a graph that shows the population of this town for the next 30 years. c. How many people live in New York in the ! year 2008?

8.872 million

d. Determine when the population will reach 10 million mathematically. _________________ " 10 % 10 t 10 t 10 10 = 7.9(1.031) = (1.031) t = 10log$ ' log1.031 = 75.145 ... year 2045 # 7.95 & 7.95 9. Exponentials and Logarithmic Functions

!

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www.mastermathmentor.com - Stu Schwartz

5. At Area Six High School with population 1,024, 18 students have colds on September 1, and the number of students with colds doubles every 11 days. Variables: po = 18

r=1

n = 11

a. Write the equation that describes this growth pattern. P = 18(1+ 1)

t

11

b.! Draw a graph that shows the number of students not having had a cold over two months. P = 1024 "18(1+ 1)

t

11

!

c. How many students did not have a cold after 30 days.

! d. When was half the student body affected?

905 students

Day 53 (Sept 1 is day 0) so October 24.

6. The population of Russia in 2006 was 142 million and because of declining births and life expectancy the growth rate is decreasing at 0.6% per year. Variables: po = 142

r = -.006

n=1

a. Write the equation that describes this decay pattern. P = 142(1" .006)

t

b.! Draw a graph that shows the population from 2006 to 2050. ! c. How many people will live in Russia in 2075?

93.745 million

d. In what year will the population be down to 100 million? Do mathematically. Verify graphically. 100 = 142(.994 ) 100 = .994 t 142

!

t

" 100 % y = log$ ' log.994 = 58.267 # 142 &

year 2064

7. The half life of marijuana varies from person to person but is normally between one to 10 days. Assume that the half-life is 5 days. Suppose a person uses marijuana today (you don’t know how much – so deal with it in terms of percent. Today he has 100% of the marijuana in his/her system. a. Write the equation that describes this decay pattern. P = 1(.5)

t

5

b. Draw a graph that shows the percent of marijuana over a month.

! 3 weeks? c. How much marijuana will be remaining after

5.4%

d. How many days will it take for less than 1% of the original marijuana to be left in that person’s bodies? 33 days

9. Exponentials and Logarithmic Functions

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www.mastermathmentor.com - Stu Schwartz

8. Complete the table. Principal a. $500 b. $2,000 c. $25,000 d. $4,000 e. $850 f. $2,400 g. $100,000 h. $100,000 i. $100,000 j. $100,000 k. $100,000 l. $100,000 m. $100,000 n. $100,000

Rate 6% 3.5% 6.2% 7.3% 3.3% 5.8% 4.5% 4.5% 4.5% 4.5% 4.5% 4.5% 4.5% 4.5%

Time 5 years 6 years 20 years 8 years 2.5 years 9 months 15 years 15 years 15 years 15 years 15 years 15 years 15 years 15 years

Method Annually Quarterly Monthly Daily Weekly Continuously Bi-annually Annually Semi-annually Quarterly Monthly Daily Hourly Continuously

New principal 691.13 2,465.10 86,114.99 7,172.37 923.07 2,506.70 190,852.89 193,528.24 194,939.34 195,664.52 196,155.50 196,395.13 196,402.96 196,403.30

9. Mr. and Mrs. B. Prepared decide at age 25 when they get married that they want to have $100,000 in the bank at the time of their retirement at age 65. They put away much of their wedding money into the bank at 5.5% interest. How much must they put in the bank, compounded daily, to have $100,000 when they retire? " .055 % 365( 40) 100000 P$1+ = 100000 P= = $11,082.15 ' 365( 40) # 365 & (1+ .055 365 ) 10. Complete the chart to find the amount of time necessary for P dollars to double if interest is compounded e rt = 2 t = lnr 2 continuously at the following rates. Pe rt = 2P ! r t

3% 23.10 yrs

4% 17.33 yrs !

5% 13.86 yrs

6% 11.55 yrs

7% 9.90 yrs

11. Complete the chart to find the amount of time necessary for P dollars to triple if interest is compounded e rt = 3 t = lnr 3 continuously at the following rates. Pe rt = 3P r t

!

3% 36.62 yrs

4% 27.47 yrs !

5% 21.97 yrs

6% 18.31 yrs

7% 15.69 yrs

12. The total interest I paid on a home mortgage of P dollars at interest rate r for t years is given by the formula: ) , + . rt + . I = P+ 12t "1. . Suppose a mortgage of $400,000 had to be paid off over 30 years at 5.5%. # & +1" % 1 ( . +* $1+ r 12 ' .a. What is the total interest and amount to be paid back over 30 years. Interest: 417,616.16 Payback 817,616.16 b. Suppose you decided to pay it back over 15 years. How do those figures change? Interest:188,300.09 Payback:588,300.09 c. Suppose you were able to get an interest rate of 5%. How does the 30 year payback figures change? Interest: 373023.14 Payback: 773,023.14 9. Exponentials and Logarithmic Functions

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www.mastermathmentor.com - Stu Schwartz

Exponential Curve Fitting For each problem, you are given a y-intercept, a point P, and one coordinate of another point Q. You are to a) determine the equation of the line that goes through the y-intercept and P, b) determine the equation of the exponential curve that goes through the y-intercept and P, and c) find the 2nd coordinate of Q in both cases. Verify both using your calculator. (Answers accurate to 2 decimal places) 1) y-intercept = 3, P (5,7), Q ( 8,_____)

2) y-intercept = 8, P (12,15), Q( 7,_____)

line work: (0,3), (5,7) 4 m= 5 4 47 ! y= x+3 y (8) = 5 5 exponential work: (0,3), (5,7) 3 = Ce 0 " C = 3 so y = 3e ax !

7 = 3e 5a " 73 = e 5a ! y (8) = 11.638

5a = ln 73 " a =

line work: (0,8), (12,15) 7 m= 12 7 145 y! = x+8 y ( 7) = 12 12 exponential work: (0,8), (12,15) 8 = Ce 0 " C = 8 so y = 8e ax 7 ln ! 3 5

15 = 8e12a " 73 = e 5a ! y ( 7) = 11.544

3) y-intercept = 25, P (6,21), Q (13, ____)

!

!

21 6a = ln 25 "a=

!

21 ln ! 25 6

5) y-intercept = 46, P (10,90), Q ( 20,_____)

!

!

line work: (0,46), (10,90) 54 m= 10 27 y!= x + 46 y (20) = 154 5 exponential work: (0,46), (10,90) 46 = Ce 0 " C = 46 so y = 46e ax 10a 90 = 46e10a " 90 46 = e ! y (20) = 176.087

line work : (0,28), (11,25) "3 m= 11 "3 y!= x + 28 y (9) = 25.546 11 exponential work : (0,28), (11,25) 28 = Ce 0 " C = 28 so y = 28e ax 11a 25 = 28e11a " 25 28 = e ! y (11) = 25.521

11a = ln 25 28 " a =

ln 25 28 11

6) y-intercept = 100, P (5,150), Q ( -3,_____)

!

line work: (0,100), (5,150)

m = 10 y = 10x + 100 !

! ln 90 46 10a = ln 90 " a = 46 10

9. Exponentials and Logarithmic Functions

!

ln 158 12

4) y-intercept = 28, P (11,15), Q ( 9,_____)

line work: (0,25), (6,21) "2 m= 3 "2 149 ! y= x + 25 y (13) = 3 3 exponential work: (0,25), (6,21) 25 = Ce 0 " C = 25 so y = 25e ax 21 21 = 25e 6a " 25 = e 6a ! y ( 7) = 17.135

12a = ln 158 " a =

exponential work: (0,100), (5,150) 100 = Ce 0 " C = 100 so y = 100e ax

150 = 100e 5a " 32 = e 5a ! y (#3) = 78.405 - 21 -

!

y ("3) = 70

5a = ln 32 " a =

ln 32 5

www.mastermathmentor.com - Stu Schwartz

7. Stopping an Oil Tanker: It has been shown that if an oil tanker stops it engines, then its velocity will decrease exponentially. Suppose that that captain shuts off the engines when the speed of the tanker is 20 miles per hour and one minute later the speed of the tanker is 18 mph. What is the speed of the tanker after one hour? How does this explain the number of tanker collisions recently in the news?

(0,20), (1,18) y = Ce at

20 = Ce 0 " C = 20 so y = 20e at

18 = 20e a " .9 = e a

a = ln.9 = #.1054

y (60) = .0356 mph 8. The Clever Rat: Psychologists have long held that much about human nature can be learned from rats. ! places a rat in a maze and that it takes the rat two minutes to escape. The second Suppose an experimenter experiment, it takes the rat 1 minute, 45 seconds to escape. How long will it take the rat to escape after the 20th trial (seconds)? How many experiments will it take the rat to get under 7 seconds?

(0,120), (1,105) y = Ce at

120 = Ce 0 " C = 120 so y = 120e at

105 = 120e a " .875 = e a

a = ln.875 = #.1335

y (19) = 9.492 sec

7 7 = 120e at " t = 120 = 21.2 trial a Answer is 20.2 trial meaning trial 21. ln

! coffee at 200o F. The teacher tells her that she may not drink it, so 9. Cooling: A student brings in a hot cup of it sits there and cools. 10 minutes later, it cooled to 140o F. How much longer will it take for the coffee to cool to 100o F? Will the decay model continue forever? If not, where does it stop?

!

( 0,200) ,(10,140)

ln0.5 = 19.434 min a The model will continue until coffee hits room temperature.

100 = 200e at " t =

y = Ce at

200 = Ce 0 " C = 200 so y = 200e at ln 0.7 140 = 200e10a " .7 = e10a a = = #.0357 10

!

10. Predicting the time of death: The Law and Order detectives enter a house at 11:30 AM and found a man ! on the floor murdered. They take his body temperature finding to be 91.8o. The body is taken to the morgue and at 12:30 PM the body temperature is 84.4o. Using an exponential model for body temperature, to the nearest minute, what time was he killed? Normal body temperature is 98.6.

(0,91.8), (60,84.4) y = Ce at

91.8 = Ce 0 " C = 91.8 so y = 91.8e at # 84.4 & ln% ( 84.4 $ 91.8 ' 60a 60a 84.4 = 91.8e " =e a= = ).0014 91.8 60

# 98.6 & ln% ( $ 91.8 ' at 98.6 = 91.8e " t = ) *51 a The person was killed at 10 : 41 AM

!

! 9. Exponentials and Logarithmic Functions

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www.mastermathmentor.com - Stu Schwartz

Use your TVM Solver to solve the following problems. 11. How much would an investment of $3,000 be worth in 10 years if it is invested at 4.5% interest compounded quarterly? $4,693.13.

!

12. Every month, I put $175 from my paycheck into an annuity at 4.75%. If I do this for a period of 5 years, how much will be in the annuity? $11.825.46.

!

13. In the situation above, I decide that I want to have $15,000 in the annuity after 5 years? How much will I have to invest a month? $221.98

!

14. I purchase a car for $35,000 and make a down payment of $5,000. So I want to finance $30,000 over 4 years. The dealer’s finance department offers me a rate of 6%. What are my monthly payments? $701.05 . How much do I actually pay back to the finance company? $33,650.40 ! !

15. In the situation above, I decide to get my own financing at a bank at 4.25%. What are my monthly payments now and how much do I pay back to the bank? $678.33,$32,559.84.

!

9. Exponentials and Logarithmic Functions

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www.mastermathmentor.com - Stu Schwartz

16. In the situation above, I decide that I cannot $678 a month so I decide to finance the car for 5 years instead of 4. What are my monthly payments now and how much do I pay back to the bank? $553.92,$33.235.20

!

17. In the above situation, I realize that I have stretched myself too far and can only afford $400 a month. I decide that 5 years is too long to finance the car so I go back to 4 years at 4.25%. How expensive a car can I afford? $17.690.49 plus the down payment of $5,000 = $22.690.49.

!

18. In the above situation, I realize that I can’t get much of a car for $22,690.49. I decided to take the payments out for 5 years again. Now how much car can I afford? $21,663.59 plus the down payment of $5,000 = $26,663.59.

!

19. In the above situation, I finally decide to finance $20,000 for 4 years and am willing to pay $450 a month. What interest rate do I need to meet these conditions? 3.99%

!

20. I purchase a house for $250,000 at 5% interest and take out a 30-year mortgage. Determine my monthly payments and the amount I pay back to the finance company. If I can reduce the interest rate to 4.75%, what will be my monthly payments and how much will I pay back to the finance company? at 5.00% ... monthly payment = $1,336.49, payback = $481,136.40 at 4.75% ... monthly payment = $1,298.98, payback = $467,632.80

!

9. Exponentials and Logarithmic Functions

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www.mastermathmentor.com - Stu Schwartz