copyright © 2013, 2009, 2005 pearson education, inc. 1 4 inverse, exponential, and logarithmic...

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

4Inverse, Exponential, and Logarithmic Functions

Copyright © 2013, 2009, 2005 Pearson Education, Inc.


4.6Applications and Models of Exponential Growth and Decay

• The Exponential Growth or Decay Function• Growth Function Models• Decay Function Models


Exponential Growth or Decay Function

In many situations that occur in ecology, biology, economics, and the social sciences, a quantity changes at a rate proportional to the amount present. In such cases the amount present at time t is a special function of t called an exponential growth or decay function.


Exponential Growth or Decay FunctionLet y0 be the amount or number present at time t = 0. Then, under certain conditions, the amount present at any time t is modeled by

0 ,kty y e

where k is a constant.


Exponential Growth or Decay Function

When k > 0, the function describes growth. In Section 4.2, we saw examples of exponential growth: compound interest and atmospheric carbon dioxide, for example.

When k < 0, the function describes decay; one example of exponential decay is radioactivity.


Example 1DETERMINING AN EXPONENTIAL FUNCTION TO MODEL THE INCREASE OF CARBON DIOXIDE

In Example 11, Section 4.2, we discussed the growth of atmospheric carbon dioxide over time. A function based on the data from the table was given in that example. Now we can see how to determine such a function from the data.

Year Carbon Dioxide (ppm)

1990 353

2000 375

2075 590

2175 1090

2275 2000



Solution

(a) Find an exponential function that gives the amount of carbon dioxide y in year x.

Recall that the graph of the data points showed exponential growth, so the equation will take the form y = y0ekt. We must find the values of y0

and k. The data began with the year 1990, so to simplify our work we let 1990 correspond to x = 0, 1991 correspond to x = 1, and so on. Since y0 is the initial amount, y0 = 353 in 1990 when x = 0.



Solution

Find an exponential function that gives the amount of carbon dioxide y in year x.

Thus the equation is

353 .kxy e

From the last pair of values in the table, we know that in 2275 the carbon dioxide level is expected to be 2000 ppm. The year 2275 corresponds to 2275 – 1990 = 285. Substitute 2000 for y and 285 for x and solve for k.

(a)



(285)2000 353 k e

2852000353

ke Divide by 353.

2852000In In

353k

e Take the logarithm on

each side.

2000In 285

353k

In ex = x, for all x.

Solution

353 xky e



1 2000In

285 353k Multiply by and

rewrite.

1285

0.00609k Use a calculator.

A function that models the data is

0.00609353 .xy e

Solution



Solution

(b)

Let y = 2(280) = 560 in y = 353e0.00609x , and find x.0.00609560 353 x e

0.00609560353

xe Divide by 353.

0.00609560In In

353x e Take the logarithm on

each side.

Estimate the year when future levels of carbon dioxide will be double the preindustrial level of 280 ppm.



560In 0.00609

353x In ex = x, for all x.

1 560In

0.00609 353x Multiply by and

rewrite.

10.00609

75.8x Use a calculator.

Since x = 0 corresponds to 1990, the preindustrial carbon dioxide level will double in the 75th year after 1990, or during 2065, according to this model.

Solution


Example 2FINDING DOUBLING TIME FOR MONEY

How long will it take for the money in an account that accrues interest at a rate of 3%, compounded continuously, to double?Solution r tA P e Continuous compounding

formula

0.032 tP P e Let A = 2P and r = 0.03.

0.032 te Divide by P.

0.03 2 In In t e Take the logarithm on each side.


Example 2FINDING DOUBLING TIME FOR MONEY

Solution In 2 0.03t

Divide by 0.03

In ex = x

In 20.03

t

23.10 t Use a calculator.

It will take about 23 yr for the amount to double.

How long will it take for the money in an account that accrues interest at a rate of 3%, compounded continuously, to double?


Example 3 DETERMINING AN EXPONENTIAL FUNCTION TO MODEL POPULATION GROWTH

According to the U.S. Census Bureau, the world population reached 6 billion people during 1999 and was growing exponentially. By the end of 2010, the population had grown to 6.947 billion. The projected world population (in billions of people) t years after 2010, is given by the function

0.00745( ) 6.947 .tt f e



0.00745( ) 6.947 tt f e

Based on this model, what will the world population be in 2020?

Solution

(a)

Since t = 0 represents the year 2010, in 2020, t would be 2020 – 2010 = 10 yr. We must find (t) when t is 10.

0.007 145( )0( ) 6.94710 f e Let t = 10.

(10) 7.484fThe population will be 7.484 billion at the end of 2020.

Given formula



0.007456.947( ) tt ef

Solution

(b)

Let f(t) = 9.

In what year will the world population reach 9 billion?

0.007459 6.947 t e

0.0074596.947

te Divide by 6.947.

0.00745In In9

6.947t e

Take the logarithm on each side.



9In 0.00745

6.947t In ex = x, for all x.

9In

6.9470.00745

t Divide by 0.00745 and rewrite.

34.8t Use a calculator.

Thus, 34.8 yr after 2010, during the year 2044, world population will reach 9 billion.

Solution

(b) In what year will the world population reach 9 billion?


Example 4DETERMINING AN EXPONENTIAL FUNCTION TO MODEL RADIOACTIVE DECAY

Suppose 600 g of a radioactive substance are present initially and 3 yr later only 300 g remain.(a) Determine the exponential equation that models this decay.Solution

( )0

0600 ky e Let y = 600 and t = 0.

0600 y e0 = 1

To express the situation as an exponential equation y = y0ekt, we use the given values first to find y0 and then to find k.

0kty y e



30.5 ke

Let y = 300 and t = 3.

3 0.5In In k e

Divide by 600.

360300 0 k e

Thus, y0 = 600, which gives y = 600ekt. Because the initial amount (600 g) decays to half that amount (300 g) in 3 yr, its half-life is 3 yr. Now we solve this exponential equation for k.

600 kty e

Take logarithms on each side.

Solution



In 0.5 3k

Divide by 3.

In ex = x, for all x.

In 0.53

k

0.231k Use a calculator.

Solution

Thus, the exponential decay equation is y = 600e –0.231t .



To find the amount present after 6 yr, let t = 6.

60.231( ) 1.386600 600 150y e e

Solution

(b) How much of the substance will be present after

6 yr?

After 6 yr, 150 g of the substance will remain.


Example 5 SOLVING A CARBON DATING PROBLEM

Carbon-14, also known as radiocarbon, is a radioactive form of carbon that is found in all living plants and animals. After a plant or animal dies, the radiocarbon disintegrates. Scientists can determine the age of the remains by comparing the amount of radiocarbon with the amount present in living plants and animals. This technique is called carbon dating. The amount of radiocarbon present after t years is given by

0.00012160 ,ty y e

where y0 is the amount present in living plants and animals.



0.00012160

tyy e

Find the half-life of carbon-14.

Solution

(a)

If y0 is the amount of radiocarbon present in a living thing, then y0 is half this initial amount. Thus, we substitute and solve the given equation for t.

0.00012100

612

ty y e Let y = y0 .1

2

1

2

Given equation



Solution

0.000121612

te Divide by y0.

0.0001216In In 12

t eTake the logarithm on each side.

Find the half-life of carbon-14.(a)

1In 0.0001 16

2 2 t In ex = x, for all x.



Solution

Divide by – 0.0001216.

1In

20.00012

16t

5700 t Use a calculator.

The half-life is about 5700 yr.

Find the half-life of carbon-14.(a)



Solution

(b) Charcoal from an ancient fire pit on Java contained ¼ the carbon-14 of a living sample of the same size. Estimate the age of the charcoal.

Solve again for t, this time letting the amount y = y0.

0.00012160

tyy e Given equation

0.00012100

614

ty y e Let y = y0 .

1

4

1

4



0.000121614

te Divide by y0 .

0.0001216In I1

n 4

t e Take the logarithm on each side.

1In

40.0001216

t

In ex = x; Divide by −0.0001216.

11,400t Use a calculator.

The charcoal is about 11,400 yr old.

Solution


Example 6 MODELING NEWTON’S LAW OF COOLING

Newton’s law of cooling says that the rate at which a body cools is proportional to the difference in temperature between the body and the environment around it. The temperature (t) of the body at time t in appropriate units after being introduced into an environment having constant temperature T0 is

0( ) ,ktt T C f e

where C and k are constants.



A pot of coffee with a temperature of 100°C is set down in a room with a temperature of 20°C. The coffee cools to 60°C after 1 hr.

Write an equation to model the data.(a)

Solution

We must find values of C and k in the given formula. From the information, when t = 0, T0 = 20, and the temperature of the coffee is f(0) = 100. Also, when t = 1, f(1) = 60. Substitute the first pair of values into the function along with T0 = 20.



0( ) ktt T C f e

Write an equation to model the data.(a)

Solution

Given formula

0100 20 kC eLet t = 0, (0) = 100, and T0 = 20.

100 20 C e0 = 1

80 C Subtract 20.

The function that models the data can now be given.

20( ) 80 ktt f e Let T0 = 20 and C = 80.



( ) 20 80 ktt f e

120 8060 k e Let t = 1 and (1) = 60.

40 80 k e Subtract 20.

12

ke Divide by 80.

1In In

2k e Take the logarithm

on each side.

Solution

Formula with T0 = 20 and C = 80.

Now use the remaining pair of values in this function to find k.



Thus, the model is 0.693( ) 20 80 .tt f e

1In

2k In ex = x for all x.

1In

2k

Solution

0.693k

Multiply by –1 and rewrite.

Use a calculator.


To find the temperature after ½ hr, let t = ½ in the model from part (a).


(b)

Solution

Model from part (a)

Let t = ½ .

Find the temperature after half an hour.

0.693( ) 20 80 tt f e

( 0.6 293)( )1/20 8012

f e

176.6 C

2 f Use a calculator.


To find how long it will take for the coffee to cool to 50°C, let (t) = 50.


(c)Solution

Let (t) = 50.

Subtract 20.

0.69320 8050 t e

How long will it take for the coffee to cool to 50°C?

0.69330 80 t e

0.69338

te Divide by 80.



Solution

How long will it take for the coffee to cool to 50°C?

0.6933In In

8t e Take the logarithm

on each side.

3In 0.693

8t In ex = x, for all x.

3In

8 1.415 hr, or about 1 hr, 25 min0.693

t

(c)

copyright © 2013, 2009, 2005 pearson education, inc. 1 4 inverse, exponential, and logarithmic...

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