8.7: Applications of Exponential Functions
Learning Objectives
 Apply the problemsolving plan to problems involving exponential functions.
 Solve realworld problems involving exponential growth.
 Solve realworld problems involving exponential decay.
Introduction
For her eighth birthday, Shelley’s grandmother gave her a full bag of candy. Shelley counted her candy and found out that there were 160 pieces in the bag. As you might suspect, Shelley loves candy, so she ate half the candy on the first day. Then her mother told her that if she eats it at that rate, the candy will only last one more day—so Shelley devised a clever plan. She will always eat half of the candy that is left in the bag each day. She thinks that this way she can eat candy every day and never run out.
How much candy does Shelley have at the end of the week? Will the candy really last forever?
Let’s make a table of values for this problem.
\begin{align*}& \text{Day} && \ \ 0 && \ 1 && \ 2 && \ 3 && \ 4 && 5 && \ 6 && \ 7\\
& \# \ \text{of candies} && 160 && 80 && 40 && 20 && 10 && 5 && 2.5 && 1.25\end{align*}
You can see that if Shelley eats half the candies each day, then by the end of the week she only has 1.25 candies left in her bag.
Let’s write an equation for this exponential function. Using the formula \begin{align*}y = A \cdot b^x\end{align*}
Now let’s graph this function. The resulting graph is shown below.
So, will Shelley’s candy last forever? We saw that by the end of the week she has 1.25 candies left, so there doesn’t seem to be much hope for that. But if you look at the graph, you’ll see that the graph never really gets to zero. Theoretically there will always be some candy left, but Shelley will be eating very tiny fractions of a candy every day after the first week!
This is a fundamental feature of an exponential decay function. Its values get smaller and smaller but never quite reach zero. In mathematics, we say that the function has an asymptote at \begin{align*}y = 0\end{align*}
ProblemSolving Strategies
Remember our problemsolving plan from earlier?
 Understand the problem.
 Devise a plan – Translate.
 Carry out the plan – Solve.
 Look – Check and Interpret.
We can use this plan to solve application problems involving exponential functions. Compound interest, loudness of sound, population increase, population decrease or radioactive decay are all applications of exponential functions. In these problems, we’ll use the methods of constructing a table and identifying a pattern to help us devise a plan for solving the problems.
Example 1
Suppose $4000 is invested at 6% interest compounded annually. How much money will there be in the bank at the end of 5 years? At the end of 20 years?
Solution
Step 1: Read the problem and summarize the information.
$4000 is invested at 6% interest compounded annually; we want to know how much money we have in five years.
Assign variables:
Let \begin{align*}x =\end{align*}
Let \begin{align*}y =\end{align*}
Step 2: Look for a pattern.
We start with $4000 and each year we add 6% interest to the amount in the bank.
\begin{align*}& \text{Start:} && \$4000\\
& 1^{st} \ \text{year:} && \text{Interest} = 4000 \times (0.06) = \$240\\
& && \text{This is added to the previous amount:} \ \$4000 + \$4000 \times (0.06)\\
& && = \$4000(1 + 0.06)\\
& && = \$4000 (1.06)\\
& && = \$4240\\
& 2^{nd} \ \text{year} && \text{Previous amount} + \text{interest on the previous amount}\\
& && = \$4240(1 + 0.06)\\
& && = \$4240 (1.06)\\
& && = \$4494.40\end{align*}
The pattern is that each year we multiply the previous amount by the factor of 1.06.
Let’s fill in a table of values:
\begin{align*}& \text{Time (years)} && \ \ 0 && \ \ 1 && \quad 2 && \quad 3 && \quad 4 && \quad 5\\
& \text{Investments amount} (\$) && 4000 && 4240 && 4494.4 && 4764.06 && 5049.90 && 5352.9\end{align*}
We see that at the end of five years we have $5352.90 in the investment account.
Step 3: Find a formula.
We were able to find the amount after 5 years just by following the pattern, but rather than follow that pattern for another 15 years, it’s easier to use it to find a general formula. Since the original investment is multiplied by 1.06 each year, we can use exponential notation. Our formula is \begin{align*}y = 4000 \cdot (1.06)^x\end{align*}
To find the amount after 5 years we plug \begin{align*}x = 5\end{align*}
\begin{align*}y = 4000 \cdot (1.06)^5 = \$5352.90\end{align*}
To find the amount after 20 years we plug \begin{align*}x = 20\end{align*}
\begin{align*}y = 4000 \cdot (1.06)^{20} = \$12828.54\end{align*}
Step 4: Check.
Looking back over the solution, we see that we obtained the answers to the questions we were asked and the answers make sense.
To check our answers, we can plug some low values of \begin{align*}x\end{align*}
\begin{align*}x = 0: \ \ y = 4000 \cdot (1.06)^0 = 4000\end{align*}
\begin{align*}x = 1: \ \ y = 4000 \cdot (1.06)^1 = 4240\end{align*}
\begin{align*}x = 2: \ \ y = 4000 \cdot (1.06)^2 = 4494.4\end{align*}
The answers match the values we found earlier. The amount of increase gets larger each year, and that makes sense because the interest is 6% of an amount that is larger every year.
Example 2
In 2002 the population of schoolchildren in a city was 90,000. This population decreases at a rate of 5% each year. What will be the population of school children in year 2010?
Solution
Step 1: Read the problem and summarize the information.
The population is 90,000; the rate of decrease is 5% each year; we want the population after 8 years.
Assign variables:
Let \begin{align*}x =\end{align*}
Let \begin{align*}y =\end{align*}
Step 2: Look for a pattern.
Let’s start in 2002, when the population is 90,000.
The rate of decrease is 5% each year, so the amount in 2003 is 90,000 minus 5% of 90,000, or 95% of 90,000.
\begin{align*}& \text{In} \ 2003: \qquad \quad \text{Population} = 90,000 \times 0.95\\
& \text{In} \ 2004: \qquad \quad \text{Population} = 90,000 \times 0.95 \times 0.95\end{align*}
The pattern is that for each year we multiply by a factor of 0.95
Let’s fill in a table of values:
\begin{align*}& \text{Year} && \ 2002 && \ 2003 && \ 2004 && \ 2005 && \ 2006 && \ 2007\\
& \text{Population} && 90,000 && 85,500 && 81,225 && 77,164 && 73,306 && 69,640\end{align*}
Step 3: Find a formula.
Since we multiply by 0.95 every year, our exponential formula is \begin{align*}y = 90000 \cdot (0.95)^x\end{align*}
\begin{align*}y = 90000 \cdot (0.95)^8 = 59,708\end{align*}
Step 4: Check.
Looking back over the solution, we see that we answered the question we were asked and that it makes sense. The answer makes sense because the numbers decrease each year as we expected. We can check that the formula is correct by plugging in the values of \begin{align*} x\end{align*}
\begin{align*}& \text{Year} \ 2002, x = 0: \qquad \quad \text{Population} = y = 90000 \cdot (0.95)^0 = 90,000\\
& \text{Year} \ 2003, x = 1: \qquad \quad \text{Population} = y = 90000 \cdot (0.95)^1 = 85,500\\
& \text{Year} \ 2004, x = 2: \qquad \quad \text{Population} = y = 90000 \cdot (0.95)^2 = 81,225\end{align*}
Solve RealWorld Problems Involving Exponential Growth
Now we’ll look at some more realworld problems involving exponential functions. We’ll start with situations involving exponential growth.
Example 3
The population of a town is estimated to increase by 15% per year. The population today is 20 thousand. Make a graph of the population function and find out what the population will be ten years from now.
Solution
First, we need to write a function that describes the population of the town.
The general form of an exponential function is \begin{align*}y = A \cdot b^x\end{align*}.
Define \begin{align*}y\end{align*} as the population of the town.
Define \begin{align*}x\end{align*} as the number of years from now.
\begin{align*}A\end{align*} is the initial population, so \begin{align*}A = 20\end{align*} (thousand).
Finally we must find what \begin{align*}b\end{align*} is. We are told that the population increases by 15% each year. To calculate percents we have to change them into decimals: 15% is equivalent to 0.15. So each year, the population increases by 15% of \begin{align*}A\end{align*}, or \begin{align*}0.15A\end{align*}.
To find the total population for the following year, we must add the current population to the increase in population. In other words, \begin{align*}A + 0.15A = 1.15A\end{align*}. So the population must be multiplied by a factor of 1.15 each year. This means that the base of the exponential is \begin{align*}b = 1.15\end{align*}.
The formula that describes this problem is \begin{align*}y = 20 \cdot (1.15)^x\end{align*}.
Now let’s make a table of values.
\begin{align*}x\end{align*}  \begin{align*}y = 20 \cdot (1.15)^x\end{align*} 

10  4.9 
5  9.9 
0  20 
5  40.2 
10  80.9 
Now we can graph the function.
Notice that we used negative values of \begin{align*}x\end{align*} in our table of values. Does it make sense to think of negative time? Yes; negative time can represent time in the past. For example, \begin{align*}x = 5\end{align*} in this problem represents the population from five years ago.
The question asked in the problem was: what will be the population of the town ten years from now? To find that number, we plug \begin{align*}x = 10\end{align*} into the equation we found: \begin{align*}y = 20 \cdot (1.15)^{10} = 80,911\end{align*}.
The town will have 80,911 people ten years from now.
Example 4
Peter earned $1500 last summer. If he deposited the money in a bank account that earns 5% interest compounded yearly, how much money will he have after five years?
Solution
This problem deals with interest which is compounded yearly. This means that each year the interest is calculated on the amount of money you have in the bank. That interest is added to the original amount and next year the interest is calculated on this new amount, so you get paid interest on the interest.
Let’s write a function that describes the amount of money in the bank.
The general form of an exponential function is \begin{align*}y = A \cdot b^x\end{align*}.
Define \begin{align*}y\end{align*} as the amount of money in the bank.
Define \begin{align*}x\end{align*} as the number of years from now.
\begin{align*}A\end{align*} is the initial amount, so \begin{align*}A = 1500\end{align*}.
Now we have to find what \begin{align*}b\end{align*} is.
We’re told that the interest is 5% each year, which is 0.05 in decimal form. When we add \begin{align*}0.05A\end{align*} to \begin{align*}A\end{align*}, we get \begin{align*}1.05A\end{align*}, so that is the factor we multiply by each year. The base of the exponential is \begin{align*}b = 1.05\end{align*}.
The formula that describes this problem is \begin{align*}y = 1500 \cdot 1.05^x\end{align*}. To find the total amount of money in the bank at the end of five years, we simply plug in \begin{align*}x = 5\end{align*}.
\begin{align*}y = 1500 \cdot (1.05)^5 = \$ 1914.42\end{align*}
Solve RealWorld Problems Involving Exponential Decay
Exponential decay problems appear in several application problems. Some examples of these are halflife problems and depreciation problems. Let’s solve an example of each of these problems.
Example 5
A radioactive substance has a halflife of one week. In other words, at the end of every week the level of radioactivity is half of its value at the beginning of the week. The initial level of radioactivity is 20 counts per second.
Draw the graph of the amount of radioactivity against time in weeks.
Find the formula that gives the radioactivity in terms of time.
Find the radioactivity left after three weeks.
Solution
Let’s start by making a table of values and then draw the graph.
Time  Radioactivity 

0  20 
1  10 
2  5 
3  2.5 
4  1.25 
5  0.625 
Exponential decay fits the general formula \begin{align*}y = A \cdot b^x\end{align*}. In this case:
\begin{align*}y\end{align*} is the amount of radioactivity
\begin{align*}x\end{align*} is the time in weeks
\begin{align*}A = 20\end{align*} is the starting amount
\begin{align*}b = \frac{1}{2}\end{align*} since the substance losses half its value each week
The formula for this problem is \begin{align*}y = 20 \cdot \left ( \frac{1}{2} \right )^x\end{align*} or \begin{align*}y = 20 \cdot 2^{x}\end{align*}. To find out how much radioactivity is left after three weeks, we plug \begin{align*}x = 3\end{align*} into this formula.
\begin{align*}y = 20 \cdot \left (\frac{1}{2} \right )^3 = 20 \cdot \left ( \frac{1}{8} \right ) = 2.5\end{align*}
Example 6
The cost of a new car is $32,000. It depreciates at a rate of 15% per year. This means that it loses 15% of each value each year.
Draw the graph of the car’s value against time in year.
Find the formula that gives the value of the car in terms of time.
Find the value of the car when it is four years old.
Solution
Let’s start by making a table of values. To fill in the values we start with 32,000 at time \begin{align*}t = 0\end{align*}. Then we multiply the value of the car by 85% for each passing year. (Since the car loses 15% of its value, that means it keeps 85% of its value). Remember that 85% means that we multiply by the decimal .85.
Time  Value (thousands) 

0  32 
1  27.2 
2  23.1 
3  19.7 
4  16.7 
5  14.2 
Now draw the graph:
Let’s start with the general formula \begin{align*}y = A \cdot b^x\end{align*}
In this case:
\begin{align*}y\end{align*} is the value of the car,
\begin{align*}x\end{align*} is the time in years,
\begin{align*}A = 32\end{align*} is the starting amount in thousands,
\begin{align*}b = 0.85\end{align*} since we multiply the amount by this factor to get the value of the car next year
The formula for this problem is \begin{align*}y = 32 \cdot (0 .85)^x\end{align*}.
Finally, to find the value of the car when it is four years old, we plug \begin{align*}x = 4\end{align*} into that formula: \begin{align*}y = 32 \cdot (0.85)^4 = 16.7\end{align*} thousand dollars, or $16,704 if we don’t round.
Review Questions
Solve the following application problems.

Halflife: Suppose a radioactive substance decays at a rate of 3.5% per hour.
 What percent of the substance is left after 6 hours?
 What percent is left after 12 hours?
 The substance is safe to handle when at least 50% of it has decayed. Make a guess as to how many hours this will take.
 Test your guess. How close were you?

Population decrease: In 1990 a rural area has 1200 bird species.
 If species of birds are becoming extinct at the rate of 1.5% per decade (ten years), how many bird species will be left in the year 2020?
 At that same rate, how many were there in 1980?
 Growth: Janine owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007?

Investment: Paul invests $360 in an account that pays 7.25% compounded annually.
 What is the total amount in the account after 12 years?
 If Paul invests an equal amount in an account that pays 5% compounded quarterly (four times a year), what will be the amount in that account after 12 years?
 Which is the better investment?
 The cost of a new ATV (allterrain vehicle) is $7200. It depreciates at 18% per year.
 Draw the graph of the vehicle’s value against time in years.
 Find the formula that gives the value of the ATV in terms of time.
 Find the value of the ATV when it is ten years old.
 A person is infected by a certain bacterial infection. When he goes to the doctor the population of bacteria is 2 million. The doctor prescribes an antibiotic that reduces the bacteria population to \begin{align*}\frac{1}{4}\end{align*} of its size each day.
 Draw the graph of the size of the bacteria population against time in days.
 Find the formula that gives the size of the bacteria population in terms of time.
 Find the size of the bacteria population ten days after the drug was first taken.
 Find the size of the bacteria population after 2 weeks (14 days).
Texas Instruments Resources
In the CK12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9618.
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