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B. Applications of Logistic, Exponential, and Logarithmic Functions

# B. Applications of Logistic, Exponential, and Logarithmic Functions

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CHAPTER 5 Exponential and Logarithmic Functions

The company originally had only a 6% market share.

M1302 ϭ

66

substitute 30 for t

1 ϩ 10eϪ0.051302

66

ϭ

1 ϩ 10eϪ1.5

Ϸ 20.4

simplify

result

After 30 days, they held a 20.4% market share.

b. For part (b), we replace M(t) with 60 and

solve for t.

66

1 ϩ 10eϪ0.05t

6011 ϩ 10eϪ0.05t 2 ϭ 66

1 ϩ 10eϪ0.05t ϭ 1.1

10eϪ0.05t ϭ 0.1

eϪ0.05t ϭ 0.01

ln eϪ0.05t ϭ ln 0.01

Ϫ0.05t ϭ ln 0.01

ln 0.01

Ϫ0.05

Ϸ 92

60 ϭ

given

multiply by 1 ϩ 10eϪ0.05t

divide by 60

subtract 1

divide by 10

b. Using the intersection-of-graphs method, we

graph Y1 with Y2 ϭ 60 to find any point(s) of

intersection. For the window size, we reason

that after 30 days, there is only a 20.4% market

share (x must be much greater than 30), and a

60% market share is being explored (y must be

greater than 60). Using the window indicated in

Figure 5.51 reveals that a 60% market share

will be attained shortly after the 92nd day.

apply base-e logarithms

Figure 5.51

Property III

80

solve for t (exact form)

approximate form

0

120

0

The company will reach a 60% market share in about 92 days.

Now try Exercises 49 and 50

P0

b to find an

P

altitude H, given a temperature and the atmospheric (barometric) pressure in centimeters of mercury (cmHg). Using the tools from this section, we are now able to find the

atmospheric pressure for a given altitude and temperature.

Earlier we used the barometric equation H ϭ 130T ϩ 80002 ln a

EXAMPLE 8

Using Logarithms to Determine Atmospheric Pressure

Suppose a group of climbers has just scaled Mt. Rainier, the highest mountain of

the Cascade Range in western Washington State. If the mountain is about 4395 m

high and the temperature at the summit is Ϫ22.5°C, what is the atmospheric

pressure at this altitude? The pressure at sea level is P0 ϭ 76 cmHg.

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Algebraic Solution

H ϭ 130T ϩ 80002 ln a

P0

b

P

given

4395 ϭ 3 301Ϫ22.52 ϩ 8000 4 ln a

4395 ϭ 7325 ln a

0.6 ϭ ln a

76

b

P

76

b

P

substitute 4395

for H, 76 for P0,

and Ϫ22.5 for T

simplify

76

b

P

divide by 7325

76

P

0.6

Pe ϭ 76

76

P ϭ 0.6

e

Ϸ 41.7

e0.6 ϭ

exponential form

multiply by P

divide by e 0.6

(exact form)

approximate form

B. You’ve just seen how

we can solve applications

involving logistic, exponential,

and logarithmic functions

535

Graphical Solution

To ensure that no algebraic errors are introduced, we’ll

enter the function as it appears after the substitutions

76

Y1 ϭ 3301Ϫ22.52 ϩ 80004 ln a b.

x

For the window size,

6000

we reason that since

x must be between 0

and 76, and y is

equal to 4395, we

0

100

only need the first

“frame” around the

window that allows a

Ϫ1000

clear view of the

intersection point. Using the window indicated shows that

at an altitude of 4395 m and a temperature of Ϫ22.5°C,

the atmospheric pressure is about 41.7 cmHg.

Now try Exercises 53 and 54 ᮣ

Additional applications involving appreciation/depreciation, Newton’s law of cooling,

space ship velocities and more, can be found in the Exercise set. See Exercises 55

through 66.

5.5 EXERCISES

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The expression log2x represents a

term,

while the expression log29 represents a

term.

2. To solve the equation ln1x ϩ 32 Ϫ ln x ϭ 7, we

like terms using logarithmic properties,

prior to writing the equation in

form.

3. If certain conditions are met, we know if

logb M ϭ logb N, then M ϭ N. This is a statement

of the

property, which is valid since

logarithmic functions are

-to.

4. Since the domain of y ϭ logb x is

. solving

logarithmic equations will sometimes produce

roots. Checking all solutions to

logarithmic equations is a necessary step.

logb 1M ϩ N2 ϭ logb 1M2 ϩ logb 1N2

logb a

logb M

M

N

logb N

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CHAPTER 5 Exponential and Logarithmic Functions

7. log 4 ϩ log1x Ϫ 72 ϭ 2

26. ln 5 ϩ ln1x Ϫ 22 ϭ 1

27. log1x ϩ 82 ϩ log x ϭ log1x ϩ 182

8. log 5 ϩ log1x Ϫ 92 ϭ 1

9. log12x Ϫ 52 Ϫ log 78 ϭ Ϫ1

10. log14 Ϫ 3x2 Ϫ log 145 ϭ Ϫ2

11. log1x Ϫ 152 Ϫ 2 ϭ Ϫlog x

12. log x Ϫ 1 ϭ Ϫlog 1x Ϫ 92

13. log 12x ϩ 12 ϭ 1 Ϫ log x

14. log 13x Ϫ 132 ϭ 2 Ϫ log x

Solve each equation using the uniqueness property.

15. log 15x ϩ 22 ϭ log 2

16. log 12x Ϫ 32 ϭ log 3

17. log4 1x ϩ 22 Ϫ log43 ϭ log4 1x Ϫ 12

18. log3 1x ϩ 62 Ϫ log3 x ϭ log3 5

19. ln 18x Ϫ 42 ϭ ln 2 ϩ ln x

20. ln 1x Ϫ 12 ϩ ln 6 ϭ ln 13x2

Solve each equation using any appropriate method.

State solutions in both exact form and in approximate

form rounded to four decimal places. Clearly identify

any extraneous roots. If there are no solutions, so state.

28. log1x ϩ 142 Ϫ log x ϭ log1x ϩ 62

29. ln12x ϩ 12 ϭ 3 ϩ ln 6

30. ln 21 ϭ 1 ϩ ln1x Ϫ 22

31. log1Ϫx Ϫ 12 ϭ log15x2 ϩ log x

32. log11 Ϫ x2 ϩ log x ϭ log1x ϩ 42

33. log1x Ϫ 12 Ϫ log x ϭ log1x Ϫ 32

34. ln x ϩ ln1x Ϫ 22 ϭ ln 4

35. 7x ϭ 231

36. 6x ϭ 3589

37. 53x Ϫ 2 ϭ 128,965

38. 93x Ϫ 3 ϭ 78,462

39. 2xϩ1 ϭ 3x

40. 7x ϭ 42xϪ1

Solve each equation using the zeroes method or the

intersection-of-graphs method. Round approximate

solutions to three decimal places.

3

41. 2

x ϭ ln1x ϩ 52

42.

x2 Ϫ 25

ϭ Ϫln1x ϩ 92 ϩ 6

x2 Ϫ 9

43. 2x

2

ϪxϪ6

ϭ x2 ϩ x Ϫ 6

21. log12x Ϫ 12 ϩ log 5 ϭ 1

1

44. x3 Ϫ 9x ϭ ex

2

22. log1x Ϫ 72 ϩ log 3 ϭ 2

45.

250

ϭ 200

1 ϩ 4eϪ0.06x

24. log3 1x Ϫ 42 ϩ log3 172 ϭ 2

46.

80

ϭ 50

1 ϩ 15eϪ0.06x

23. log2 192 ϩ log2 1x ϩ 32 ϭ 3

25. ln1x ϩ 72 ϩ ln 9 ϭ 2

WORKING WITH FORMULAS

47. Logistic growth: P1t2 ‫؍‬

C

1 ؉ ae؊kt

For populations that exhibit logistic growth, the

population at time t is modeled by the function

shown, where C is the carrying capacity of the

population (the maximum population that can be

supported over a long period of time), k is the

P102

growth constant, and a ϭ C Ϫ

P102 . Solve the

formula for t, then use the result to find the value of

t given C ϭ 450, a ϭ 8, P ϭ 400, and k ϭ 0.075.

48. Estimating time of death: h ‫ ؍‬؊3.9 # lna

T ؊ TR

b

T0 ؊ TR

Using the formula shown, a forensic expert can

compute the approximate time of death for a

person found recently expired, where T is the body

temperature when it was found, TR is the (constant)

temperature of the room, T0 is the body

temperature at the time of death (T0 ϭ 98.6°F), and

h is the number of hours since death. If the body

was discovered at 9:00 A.M. with a temperature of

86.2°F, in a room at 73°F, at approximately what

time did the person expire? (Note this formula is a

version of Newton’s law of cooling.)

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APPLICATIONS

Use the barometric equation H ‫ ؍‬130T ؉ 80002 ln a

P0

b

P

for exercises 49 and 50. Recall that P0 ‫ ؍‬76 cmHg.

49. Altitude and temperature: A sophisticated spy

plane is cruising at an altitude of 18,250 m. If the

temperature at this altitude is Ϫ75°C, what is the

barometric pressure?

50. Altitude and temperature: A large weather

balloon is released and takes altitude, pressure, and

information back to Earth. What is the pressure

reading at an altitude of 5000 m, given the

temperature is Ϫ18°C?

51. Stocking a lake: A farmer wants to stock a private

lake on his property with catfish. A specialist studies

the area and depth of the lake, along with other

factors, and determines it can support a maximum

population of around 750 fish, with growth modeled

750

by the function P1t2 ϭ

, where P(t)

1 ϩ 24eϪ0.075t

gives the current population after t months. (a) How

many catfish did the farmer initially put in the

lake? (b) How many months until the population

reaches 300 fish?

52. Increasing sales: After expanding their area of

operations, a manufacturer of small storage

buildings believes the larger area can support sales

of 40 units per month. After increasing the

advertising budget and enlarging the sales force,

sales are expected to grow according to the model

40

S1t2 ϭ

, where S(t) is the expected

1 ϩ 1.5eϪ0.08t

number of sales after t months. (a) How many sales

were being made each month, prior to the expansion?

(b) How many months until sales reach 25 units per

month?

Use Newton’s law of cooling T ‫ ؍‬TR ؉ (T0 ؊ TR)ekh

to complete Exercises 57 and 58. Recall that water

freezes at 32؇F and use k ‫ ؍‬؊0.012. Refer to

Section 5.2, page 498 as needed.

53. Making popsicles: On a hot summer day, Sean

and his friends mix some Kool-Aid® and decide to

freeze it in an ice tray to make popsicles. If the

water used for the Kool-Aid® was 75°F and the

freezer has a temperature of Ϫ20°F, how long will

they have to wait to enjoy the treat?

54. Freezing time: Suppose the current temperature in

Esconabe, Michigan, was 47°F when a 5°F arctic

cold front moved over the state. How long would it

take a puddle of water to freeze over?

Depreciation/appreciation: As time passes, the value of

certain items decrease (appliances, automobiles, etc.),

while the value of other items increase (collectibles,

real estate, etc.). The time T in years for an item to

reach a future value can be modeled by the formula

Vn

T ‫ ؍‬k ln a b, where Vn is the purchase price when

Vf

new, Vf is its future value, and k is a constant that

depends on the item.

55. Automobile depreciation: If a new car is purchased

for \$28,500, find its value 3 yr later if k ϭ 5.

56. Home appreciation: If a new home in an

“upscale” neighborhood is purchased for \$130,000,

find its value 12 yr later if k ϭ Ϫ16.

Drug absorption: The time required for a certain

percentage of a drug to be absorbed by the body after

injection depends on the drug’s absorption rate. This

؊ln p

can be modeled by the function T( p) ‫؍‬

, where

k

p represents the percent of the drug that remains

unabsorbed (expressed as a decimal), k is the absorption

rate of the drug, and T( p) represents the elapsed time.

57. For a drug with an absorption rate of 7.2%, (a) find

the time required (to the nearest hour) for the body

to absorb 35% of the drug, and (b) find the percent

of this drug (to the nearest half percent) that

remains unabsorbed after 24 hr.

58. For a drug with an absorption rate of 5.7%, (a) find

the time required (to the nearest hour) for the body

to absorb 50% of the drug, and (b) find the percent

of this drug (to the nearest half percent) that

remains unabsorbed after 24 hr.

Spaceship velocity: In space travel, the change in the

velocity of a spaceship Vs (in km/sec) depends on the

mass of the ship Ms (in

tons), the mass of the fuel

which has been burned Mf

(in tons) and the escape

velocity of the exhaust Ve

(in km/sec). Disregarding

frictional forces, these are

related by the equation

Ms

b.

Vs ‫ ؍‬Ve ln a

Ms Ϫ Mf

59. For the Jupiter VII rocket, find the mass of the fuel

Mf that has been burned if Vs ϭ 6 km/sec when

Ve ϭ 8 km/sec, and the ship’s mass is 100 tons.

60. For the Neptune X satellite booster, find the mass

of the ship Ms if Mf ϭ 75 tons of fuel has been

burned when Vs ϭ 8 km/sec and Ve ϭ 10 km/sec.

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Learning curve: The job performance of a new

employee when learning a repetitive task (as on an

assembly line) improves very quickly at first, then

grows more slowly over time. This can be modeled by

the function P(t) ‫ ؍‬a ؉ b ln t, where a and b are

constants that depend on the type of task and the

training of the employee.

61. The number of toy planes an employee can

assemble from its component parts depends on the

length of time the employee has been working.

This output is modeled by P1t2 ϭ 5.9 ϩ 12.6 ln t,

where P(t) is the number of planes assembled daily

after working t days. (a) How many planes is an

employee making after 5 days on the job? (b) How

many days until the employee is able to assemble

34 planes per day?

62. The number of circuit boards an associate can

assemble from its component parts depends on the

length of time the associate has been working. This

output is modeled by B1t2 ϭ 1 ϩ 2.3 ln t, where

B(t) is the number of boards assembled daily after

working t days. (a) How many boards is an

employee completing after 9 days on the job?

(b) How long will it take until the employee is able

to complete 10 boards per day?

EXTENDING THE CONCEPT

Solve the following equations. Note that equations Exercises 63 and 64 are in quadratic form.

63. 2e2x Ϫ 7ex ϭ 15

64. 3e2x Ϫ 4ex Ϫ 7 ϭ Ϫ3

65. Use the algebraic method to find the inverse

function.

a. f 1x2 ϭ 2xϩ1

b. y ϭ 2 ln 1x Ϫ 32

66. Show that g1x2 ϭ f Ϫ1 1x2 by composing the

functions.

a. f 1x2 ϭ 3xϪ2; g1x2 ϭ log3 x ϩ 2

b. f 1x2 ϭ exϪ1; g1x2 ϭ ln x ϩ 1

67. Use properties of logarithms and/or exponents to

show

a. y ϭ 2x is equivalent to y ϭ ex ln 2.

b. y ϭ bx is equivalent to y ϭ erx,

where r ϭ ln b.

68. Use test values for p and q to demonstrate that the

following relationships are false.

a. ln 1pq2 ϭ ln p ln q b. ln p ϩ ln q ϭ ln1p ϩ q2

p

ln p

c. ln a b ϭ

q

ln q

69. Match each equation with the most appropriate solution strategy, and justify/discuss why.

a. exϩ1 ϭ 25

apply base-10 logarithm to both sides

b. log12x ϩ 32 ϭ log 53

rewrite and apply uniqueness property for exponentials

c. log1x2 Ϫ 3x2 ϭ 2

apply uniqueness property for logarithms

2x

d. 10 ϭ 97

apply either base-10 or base-e logarithm

5xϪ3

ϭ 32

e. 2

apply base-e logarithm

xϩ2

ϭ 23

f. 7

write in exponential form

70. (3.3) Match the graph shown

with its correct equation,

without actually graphing

the function.

a. y ϭ x2 ϩ 4x Ϫ 5

b. y ϭ Ϫx2 Ϫ 4x ϩ 5

c. y ϭ Ϫx2 ϩ 4x ϩ 5

d. y ϭ x2 Ϫ 4x Ϫ 5

72. (4.5) Graph the function r 1x2 ϭ

y

10

x2 Ϫ 4

. Label all

xϪ1

intercepts and asymptotes.

Ϫ10

10 x

Ϫ10

71. (2.3/2.4) State the domain and range of the

functions.

a. y ϭ 12x ϩ 3

b. y ϭ Ϳx ϩ 2Ϳ Ϫ 3

73. (2.6) Suppose the maximum load (in tons) that can

be supported by a cylindrical post varies directly

with its diameter raised to the fourth power and

inversely as the square of its height. A post 8 ft

high and 2 ft in diameter can support 6 tons. How

many tons can be supported by a post 12 ft high

and 3 ft in diameter?

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5.6

Applications from Business, Finance, and Science

LEARNING OBJECTIVES

In Section 5.6 you will see

how we can:

A. Calculate simple interest

and compound interest

Would you pay \$750,000 for a home worth only \$250,000? Surprisingly, when a conventional mortgage is repaid over 30 years, this is not at all rare. Over time, the accumulated interest on the mortgage is easily more than two or three times the original

value of the house. In this section we explore how interest is paid or charged, and look

at other applications of exponential and logarithmic functions from business, finance,

as well as the physical and social sciences.

B. Calculate interest

compounded continuously

C. Solve applications

of annuities and

amortization

D. Solve applications of

exponential growth and

decay

WORTHY OF NOTE

A. Simple and Compound Interest

Simple interest is an amount of interest that is computed only once during the lifetime

of an investment (or loan). In the world of finance, the initial deposit or base amount is

referred to as the principal p, the interest rate r is given as a percentage and stated as

an annual rate, with the term of the investment or loan most often given as time t in

years. Simple interest is merely an application of the basic percent equation, with the

additional element of time coming into play: interest ϭ principal ϫ rate ϫ time, or

I ϭ prt. To find the total amount A that has accumulated (for deposits) or is due (for

loans) after t years, we merely add the accumulated interest to the initial principal:

A ϭ p ϩ prt.

Simple Interest Formula

If a loan is kept for only a certain

number of months, weeks, or days,

the time t should be stated as a

fractional part of a year so the time

period for the rate (years) matches

the time period over which the loan

is repaid.

If principal p is deposited or borrowed at interest rate r for a period of t years, the

simple interest on this account will be

I ϭ prt

The total amount A accumulated or due after this period will be

A ϭ p ϩ prt or A ϭ p11 ϩ rt2

EXAMPLE 1

Solving an Application of Simple Interest

Many finance companies offer what have become known as PayDay Loans — a

small \$50 loan to help people get by until payday, usually no longer than 2 weeks.

If the cost of this service is \$12.50, determine the annual rate of interest charged by

these companies.

Solution

The interest charge is \$12.50, the initial principal is \$50.00, and the time period is

2

1

ϭ 26

2 weeks or 52

of a year. The simple interest formula yields

I ϭ prt

12.50 ϭ 50r a

6.5 ϭ r

simple interest formula

1

b

26

1

for t

substitute \$12.50 for I, \$50.00 for p, and 26

solve for r

The annual interest rate on these loans is a whopping 650%!

Now try Exercises 7 through 16

5–61

539

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Compound Interest

Many financial institutions pay compound interest on deposits they receive, which is

interest paid on previously accumulated interest. The most common compounding periods are yearly, semiannually (two times per year), quarterly (four times per year),

monthly (12 times per year), and daily (365 times per year). Applications of compound

interest typically involve exponential functions. For convenience, consider \$1000 in principal, deposited at 8% for 3 yr. The simple interest calculation shows \$240 in interest is

earned and there will be \$1240 in the account: A ϭ 1000 31 ϩ 10.082132 4 ϭ \$1240. If

the interest is compounded each year 1t ϭ 12 instead of once at the start of the 3-yr period,

the interest calculation shows

A1 ϭ 100011 ϩ 0.082 ϭ 1080 in the account at the end of year 1,

A2 ϭ 108011 ϩ 0.082 ϭ 1166.40 in the account at the end of year 2,

A3 ϭ 1166.4011 ϩ 0.082 Ϸ 1259.71 in the account at the end of year 3.

The account has earned an additional \$19.71 interest. More importantly, notice

that we’re multiplying by 11 ϩ 0.082 each compounding period, meaning results can

be computed more efficiently by simply applying the factor 11 ϩ 0.082 t to the initial

principal p. For example,

A3 ϭ 100011 ϩ 0.082 3 Ϸ \$1259.71.

In general, for interest compounded yearly the accumulated value is

A ϭ p11 ϩ r2 t. Notice that solving this equation for p will tell us the amount we need

A

to deposit now, in order to accumulate A dollars in t years: p ϭ 11 ϩ

r2 t . This is called

the present value equation.

Interest Compounded Annually

If a principal p is deposited at interest rate r and compounded yearly for a period of

t yr, the accumulated value is

A ϭ p11 ϩ r2 t

If an accumulated value A is desired after t yr, and the money is deposited at interest

rate r and compounded yearly, the present value is

EXAMPLE 2

A

11 ϩ r2 t

Finding the Doubling Time for Interest Compounded Yearly

An initial deposit of \$1000 is made into an account paying 6% compounded yearly.

How long will it take for the money to double?

Solution

Using the formula for interest compounded yearly we have

A ϭ p11 ϩ r2 t

2000 ϭ 100011 ϩ 0.062 t

2 ϭ 1.06 t

ln 2 ϭ t ln 1.06

ln 2

ϭt

ln 1.06

11.9 Ϸ t

given

substitute 2000 for A, 1000 for p, and 0.06 for r

isolate variable term

apply base-e logarithms; power property

solve for t

approximate form

The money will double in just under 12 yr.

Now try Exercises 17 through 22

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B. Applications of Logistic, Exponential, and Logarithmic Functions

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