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Applications of Exponential Functions

Word problems with variables as exponents

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Exponential Growth and Decay MCC9-12.F.LE.1c

Have you ever heard of the Consumer Price Index? It measures the price level of certain goods and services in order to measure inflation. Suppose the Consumer Price Index is currently at 226 and is increasing at a rate of 3% per year. What will it be in 10 years? What exponential function could you set up to answer this question? In this Concept, you'll learn to solve real-world problems like this one by using everything that was presented in the previous Concepts.

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Multimedia Link: To learn more about how to use the correct exponential function, visit the http://regentsprep.org/REgents/math/ALGEBRA/AE7/ExpDecayL.htm - algebra lesson page by RegentsPrep.


We have to deal with problem solving in many real-world situations. Therefore, it is important to know the steps you must take when problem solving.

Example A

Suppose $4000 is invested at a 6% interest rate compounded annually. How much money will there be in the bank at the end of five years? At the end of 20 years?


Read the problem and summarize the information.

$4000 is invested at a 6% interest rate compounded annually. We want to know how much money we will have after five years.

  • Assign variables. Let \begin{align*}x=\end{align*} time in years and \begin{align*}y=\end{align*} amount of money in investment account.
  • We start with $4000 and each year we apply a 6% interest rate on the amount in the bank.
  • The pattern is that each year we multiply the previous amount by a factor of \begin{align*}100\%+6\%=106\%=1.06\end{align*}.
  • Complete a table of values.
Time (years) 0 1 2 3 4 5
Investment Amount ($) 4000 4240 4494.40 4764.06 5049.91 5352.90

Using the table, we see that at the end of five years we have $5352.90 in the investment account.

In the case of five years, we don’t need an equation to solve the problem. However, if we want the amount at the end of 20 years, it becomes too difficult to constantly multiply. We can use a formula instead.

Since we take the original investment and keep multiplying by the same factor of 1.06, this means we can use exponential notation.

\begin{align*}y=4000 \cdot (1.06)^x\end{align*}

To find the amount after five years we use \begin{align*}x=5\end{align*} in the equation.

\begin{align*}y=4000 \cdot (1.06)^5=\$ 5352.90\end{align*}

To find the amount after 20 years we use \begin{align*}x=20\end{align*} in the equation.

\begin{align*}y=4000 \cdot (1.06)^{20}=\$ 12,828.54\end{align*}

To check our answers we can plug in some low values of \begin{align*}x\end{align*} to see if they match the values in the table:

\begin{align*}& x = 0 && 4000 \cdot (1.06)0 = 4000\\ & x = 1 && 4000 \cdot (1.06)1 = 4240\\ & x = 2 && 4000 \cdot (1.06)2 = 4494.40\end{align*}

The answers make sense because after the first year, the amount goes up by $240 (6% of $4000). The amount of increase gets larger each year and that makes sense because the interest is 6% of an amount that is larger and larger every year.

Example B

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 its value each year.

  • Draw a graph of the car’s value against time in years.
  • 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.


This is an example of an exponential decay function. Start by making a table of values. To fill in the values we start with 32,000 when \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, it keeps 85% of its value). Remember \begin{align*}85\% =0.85\end{align*}.

Time Value (Thousands)
0 32
1 27.2
2 23.1
3 19.7
4 16.7
5 14.2

The general formula is \begin{align*}y=a (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=32,000\end{align*} is the starting amount in thousands, and \begin{align*}b=0.85\end{align*} since we multiply the value in any year by this factor to get the value of the car in the following year. The formula for this problem is \begin{align*}y=32,000 (0.85)^x\end{align*}.

Finally, to find the value of the car when it is four years old, we use \begin{align*}x=4\end{align*} in the formula. Remember the value is in thousands.


Example C

The half-life of the prescription medication Amiodarone is 25 days. Suppose a patient has a single dose of 12 mg of this drug in her system.

  1. How much Amiodarone will be in the patient’s system after four half-life periods?
  2. When will she have less than 3 mg of the drug in her system?


1. Four half life periods means the drug reduces by half 4 times:


Since the patient started with 12 mg, \begin{align*} 12\cdot \frac{1}{16}=\frac{3\cdot 4}{4\cdot 4}=\frac{3}{4}\end{align*}.

There will be \begin{align*}\frac{3}{4}\end{align*} mg of Amiodarone left in the patient's system after 4 half lives.

2. 3 is half of 6, which is half of 12. So 12 mg will be reduced to 3 mg after two half lives. Thus, after 50 days, the patient will have less than 3 mg of Amiodarone in his or her system.

Guided Practice

The population of a town is estimated to increase by 15% per year. The population today is 20,000. Make a graph of the population function and find out what the population will be ten years from now.

Solution: The population is growing at a rate of 15% each year. When something grows at a percent, this is a clue to use exponential functions.

Remember, the general form of an exponential function is \begin{align*}y=a(b)^x\end{align*}, where \begin{align*}a\end{align*} is the beginning value and \begin{align*}b\end{align*} is the total growth rate. The beginning value is 20,000. Therefore, \begin{align*}a=20,000\end{align*}.

The population is keeping the original number of people and adding 15% more each year.


Therefore, the population is growing at a rate of 115% each year. Thus, \begin{align*}b=1.15\end{align*}.

The function to represent this situation is \begin{align*}y=20,000 \ (1.15)^x\end{align*}.

Now make a table of values and graph the function.

\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

Notice that we used negative values of \begin{align*}x\end{align*} in our table. Does it make sense to think of negative time? In this case \begin{align*}x=-5\end{align*} represents what the population was five years ago, so it can be useful information.

The question asked in the problem was “What will be the population of the town ten years from now?” To find the population exactly, we use \begin{align*}x=10\end{align*} in the formula. We find \begin{align*}y=20,000 \cdot (1.15)^{10}=80,912\end{align*}.


Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Word Problem Solving (7:21)

Apply the problem-solving techniques described in this section to solve the following problems.

  1. Half-life Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after six hours?
  2. Population decrease In 1990, a rural area had 1200 bird species. If species of birds are becoming extinct at the rate of 1.5% per decade (10 years), how many bird species will there be left in the year 2020?
  3. Growth Nadia owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores did the restaurant operate in 2007?
  4. Investment Peter invests $360 in an account that pays 7.25% compounded annually. What is the total amount in the account after 12 years?
  5. Solve the following problems.
  6. The population of a town in 2007 is 113,505 and is increasing at a rate of 1.2% per year. What will the population be in 2012?
  7. A set of bacteria begins with 20 and doubles every 2 hours. How many bacteria would be present 15 hours after the experiment began?
  8. The cost of manufactured goods is rising at the rate of inflation, or at about 2.3%. Suppose an item costs $12 today. How much will it cost five years from now due to inflation?

Solve the following application problems.

  1. The cost of a new ATV (all-terrain vehicle) is $7200. It depreciates at 18% per year.
    1. Draw the graph of the vehicle’s value against time in years.
    2. Find the formula that gives the value of the ATV in terms of time.
    3. Find the value of the ATV when it is ten years old.
  1. Michigan’s population is declining at a rate of 0.5% per year. In 2004, the state had a population of 10,112,620.
    1. Write a function to express this situation.
    2. If this rate continues, what will the population be in 2012?
    3. When will the population of Michigan reach 9,900,000?
    4. What was the population in the year 2000, according to this model?
  1. A certain radioactive substance has a half-life of 27 years. An organism contains 35 grams of this substance on day zero.
    1. Draw the graph of the amount remaining. Use these values for \begin{align*}x: x=0, 27, 54, 81, 108, 135.\end{align*}
    2. Find the function that describes the amount of this substance remaining after \begin{align*}x\end{align*} days.
    3. Find the amount of radioactive substance after 92 days.

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