# Exponential and Logarithmic Models

#### Scatter Plot

Contents: This page corresponds to § 4.5 (p. 359) of the text.

Suggested problems from text:

p. 366 #7,9,15,17,25,33,35,69

## Exponential Growth and Decay

Two common types of mathematical models are

Exponential Growth: y = a e bx, b > 0.

Exponential Decay: y = a e -bx, b > 0.

Example 1.

During the 1980s the population of a certain city went from 100,000 to 205,000. Populations by year are listed in the table below.

 Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 Population in thousands 100 108 117 127 138 149 162 175 190 205

This data is approximated well by the exponential growth model P = 100 e0.08t, where t is the number of years since 1980. In other words, the year 1980 corresponds to t = 0, 1981 corresponds to t = 1, etc. The data points and model are graphed below.

Population data points and model P = 100 e0.08t

where t is number of years since 1980.

Problem 1: Use the model to predict the population of the city in 1994.

1991 corresponds to t = 11, so our model predicts that the population will be

P = 100 e0.08*11 = 241 thousand.

Problem 2: According to our model, when will the population reach 300 thousand?

To solve this problem we set 100 e0.08t equal to 300 and solve for t.

100 e0.08t = 300

e0.08t = 3 Take the natural logarithm of both sides.

ln e0.08t = ln 3

0.08t = ln 3

t = (ln 3)/0.08 = 13.73, approximately.

Therefore, the population is expected to reach 300 thousand about three fourths of the way through the year 1993.

It is important to recognize the limitations of this model. While it is obvious from the graph that for t between 0 and 9, the model values are very close to the actual population values, we should not assume that our model will give an accurate prediction of population for values of t much larger than 9. For instance, the model predicts that in the year 2080 (t = 100), the population of the city will be almost 300 million! That is not likely.

Suppose we know that a variable y can be expressed in the form aebx, but we don't know the values a and b. If we are given any two points on the graph of y, then it is possible to find the numbers a and b. The simplest case, and one that is often encountered in applications, is where we know the value of y when x = 0 and one other point on the graph of y.

Example 2.

Assume that a population P is growing exponentially, so P = aebt, where t is measured in years.

If P = 15000 in 1990, and P has grown to 17000 in 1993, find the formula for P.

Let t be the number of years since 1990. Then a = 15000, the value of P when t = 0.

Note that t could have been chosen differently. For instance, we could let t be the number of years since 1900. But then we would not know the value of P when t = 0, so we would not know the value of a immediately.

We still need to find b. All we know about b is that it is positive, since the population is growing. Using the value we have found for a, we have

P = 15000 ebt.

The year 1993 corresponds to t = 3, so we substitute P = 17000 and t = 3 in the equation above and solve for b.

17000 = 15000 e3b

We have solved equations of this form several times. The first step is to isolate the exponential term. Then take the natural logarithm of both sides.

17/15 = e3b

ln (17/15) = ln e3b

ln (17/15) = 3b

b = (ln(17/15))/3 = 0.0417 (approximately)

Therefore P = 15000 e0.0417 t

Based on this model, when will the population reach 20000?

Set P equal to 20000 and solve for t.

20000 = 15000 e0.0417 t

4/3 = e0.0417 t

ln (4/3) = ln e0.0417 t

ln (4/3) = 0.0417 t

t = (ln (4/3))/0.0417 = 6.9 years

So, we expect the population to reach 20000 toward the end of 1996.

Exercise 1:

Find a model of the type P = aebt, where t is the number of years since 1970, if P = 30000 in 1970 and P = 36000 in 1977. Use this model to predict the value of P in 1980.

When the coefficient of x (or whatever the independent variable is named) is negative, then we are modeling a decreasing variable. This is called exponential decay. We will illustrate exponential decay by considering a radioactive substance. A sample of a radioactive substance decays with time.

Example 3.

The mass (in grams) of radioactive material in a sample is given by N = 100e-0.0017t, where t is measured in years.

Find the half-life of this radioactive substance.

The half-life of a radioactive substance is the amount of time required for half of a give sample to decay. Note that half-life is independent of the size of the sample. If the half-life of a certain radioactive material is 700 years, then if the initial mass of the sample is 1000 grams, in 700 years there will be 500 grams. If the initial mass of the sample is only 8 grams, in 700 yeas there will be 4 grams.

In this example, the mass of the radioactive material is 100 grams at time t = 0. Therefore the half-life is the amount of time necessary for the sample to decay to 50 grams. So we can find the half-life by setting N equal to 50 and solving for t.

100e-0.0017t = 50

e-0.0017t = 0.5

Note: If the initial amount had been 800 grams, then we would have to solve the equation 800e-0.0017t=400, and after dividing both sides by 800 we would have e-0.0017t = 0.5, which is the same as the equation above. That is why half-life is independent of initial quantity.

ln e-0.0017t = ln 0.5

-0.0017t = ln 0.5

t = (ln 0.5)/ -0.0017 = 408 years (approximately)

Question: Why is (ln 0.5)/-0.0017 equal to (ln 2)/0.0017 ?

In an earlier section we discussed an important example of exponential growth, namely continuous compounding of interest. Computing the half-life of a radioactive substance is very similar to computing the doubling time for an investment.

Example 4.

If \$1000 is invested at 9% annual interest compounded continuously, how long will it take for the investment to double? Using the compound interest formula A = Pert, we have

A = 1000e0.09t

We want to find the amount of time it will take for the investment to double, that is grow to \$2000, so we set A equal to 2000 and solve for t.

1000e0.09t = 2000 Divide both sides by 1000.

e0.09t = 2

Note: Just like half-life, doubling time is independent of the initial investment P. If we had started with \$50 and asked how long it will be before we have \$100, then we would have solved the equation

50e0.09t = 100, and after dividing both sides by 50 we would again have e0.09t = 2.

ln e0.09t = ln 2

0.09 t = ln 2

t = (ln 2)/0.09 = 7.7 years (approximately).

Question: If you invest \$3000 at 9% compounded continuously, about how much will you have in 15 and a half years?

Answer: Around \$12000. This is easy to approximate because doubling time is 7.7 years, so 15.5 years corresponds to just over two doublings. 3000 -> 6000 -> 12000.

Exercise 2:

How long does it take an investment to triple at %11 compounded continuously? Answer

## Other Models

Example 5.

Consider the following data points.

 x 1 2 3 4 5 6 7 y 0.5 2.6 3.6 4.5 4.9 5.4 5.7

When we plot these points we see that they do not seem to lie on any exponential curve, but the shape is very much like a logarithmic graph. This suggests that a logarithmic model is reasonable. The graph below shows the data points and the function y = 0.63 + 2.7 ln x which fits the data points quite well.