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# 8.6: Exponential Decay Functions

Created by: CK-12

## Learning Objectives

• Graph an exponential decay function.
• Compare graphs of exponential decay functions.
• Solve real-world problems involving exponential decay.

## Introduction

In the last section, we looked at graphs of exponential functions. We saw that exponentials functions describe a quantity that doubles, triples, quadruples, or simply gets multiplied by the same factor. All the functions we looked at in the last section were exponentially increasing functions. They started small and then became large very fast. In this section, we are going to look at exponentially decreasing functions. An example of such a function is a quantity that gets decreased by one half each time. Let’s look at a specific example.

For her fifth birthday, Nadia’s grandmother gave her a full bag of candy. Nadia counted her candy and found out that there were 160 pieces in the bag. As you might suspect Nadia loves candy so she ate half the candy on the first day. Her mother told her that if she eats it at that rate it will be all gone the next day and she will not have anymore until her next birthday. Nadia devised a clever plan. She will always eat half of the candy that is left in the bag each day. She thinks that she will get candy every day and her candy will never run out. How much candy does Nadia have at the end of the week? Would the candy really last forever?

Let’s make a table of values for this problem.

$& \text{Day} && 0 && 1 && 2 && 3 && 4 && 5 && 6 && 7 \\& \text{No. of Candies} && 160 && 80 && 40 && 20 && 10 && 5 && 2.5 && 1.25$

You can see that if Nadia 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.

$& \text{Nadia started with}\ 160\ \text{pieces}. && y = 160 \\& \text{After the first she has}\ \frac{1}{2}\ \text{of that amount.} && y = 160 \cdot \frac{1} {2} \\& \text{After the second day she has}\ \frac{1}{2}\ \text{of the last amount. } && y = 160 \cdot \frac{1} {2}\cdot \frac{1} {2}$

You see that in order to get the amount of candy left at the end of each day we keep multiplying by $\frac{1}{2}$.

We can write the exponential function as

$y = 160 \cdot \frac{1} {2}^x$

Notice that this is the same general form as the exponential functions in the last section.

$y=A \cdot b^x$

Here $A = 160$ is the initial amount and $b =\frac{1}{2}$ is the factor that the quantity gets multiplied by each time. The difference is that now $b$ is a fraction that is less than one, instead of a number that is greater than one.

This is a good rule to remember for exponential functions.

If $b$ is greater than one, then the exponential function increased, but

If $b$ is less than one (but still positive), then the exponential function decreased

Let’s now graph the candy problem function. The resulting graph is shown below.

So, will Nadia’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 will see that the graph never really gets to zero.

Theoretically there will always be some candy left, but she 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 value get smaller and smaller and approaches zero but it never quite gets there. In mathematics we say that the function asymptotes to the value zero. This means that it approaches that value closer and closer without ever actually getting there.

## Graph an Exponential Decay Function

The graph of an exponential decay function will always take the same basic shape as the one in the previous figure. Let’s graph another example by making a table of values.

Example 1

Graph the exponential function $y = 5 \cdot \left (\frac{1} {2} \right)^x$

Solution

Let’s start by making a table of values.

$x$ $y = 5 \cdot \left (\frac{1} {2} \right)^x$
$-3$ $y = 5 \cdot \left (\frac{1} {2} \right)^{-3} = 5.2^3 = 40$
-2 $y = 5 \cdot \left (\frac{1} {2} \right)^{-2} = 5.2^2 = 20$
-1 $y = 5 \cdot \left (\frac{1} {2} \right)^{-1} = 5.2^1 = 10$
0 $y = 5 \cdot \left (\frac{1} {2} \right)^0 = 5.1 = 5$
1 $y = 5 \cdot \left (\frac{1} {2} \right)^1 = \frac{5} {2}$
2 $y = 5 \cdot \left (\frac{1} {2} \right)^2 = \frac{5} {4}$

Now let's graph the function.

Remember that a fraction to a negative power is equivalent to its reciprocal to the same positive power.

We said that an exponential decay function has the same general form as an exponentially increasing function, but that the base $b$ is a positive number less than one. When $b$ can be written as a fraction, we can use the Property of Negative Exponents that we discussed in Section 8.3 to write the function in a different form.

For instance, $y = 5 \cdot \left (\frac{1} {2} \right) ^x$ is equivalent to $5 \cdot 2 ^{-x}$.

These two forms are both commonly used so it is important to know that they are equivalent.

Example 2

Graph the exponential function $y=8\cdot 3^{-x}$.

Solution

Here is our table of values and the graph of the function.

$x$ $y =8\cdot 3^{-x}$
$-3$ $y=8.3^{-(-3)}=8\cdot 3^{3}= 216$
-2 $y =8.3^{-(-2)}=8\cdot 3^{2}=72$
-1 $y=8.3^{-(-1)}=8\cdot 3^1=24$
0 $y=8\cdot 3^0=8$
1 $y=8 \cdot 3^{-1}=\frac{8}{3}$
2 $y=8\cdot 3^{-2}=\frac{8}{9}$

## Compare Graphs of Exponential Decay Functions

You might have noticed that an exponentially decaying function is very similar to an exponentially increasing function. The two types of functions behave similarly, but they are backwards from each other.

The increasing function starts very small and increases very quickly and ends up very, very big. While the decreasing function starts very big and decreases very quickly to soon become very, very small. Let’s graph two such functions together on the same graph and compare them.

Example 3

Graph the functions $y=4^x$ and $y=4^{-x}$ on the same coordinate axes.

Solution

Here is the table of values and the graph of the two functions.

Looking at the values in the table we see that the two functions are “backwards” of each other in the sense that the values for the two functions are reciprocals.

$x$ $y =4^x$ $y=4^{-x}$
$-3$ $y =4^{-3} = \frac{1}{64}$ $y=4^{-(-3)} = 64$
-2 $y =4^{-2} = \frac{1}{16}$ $y=4^{-(-2)} = 16$
-1 $y =4^{-1}=\frac{1}{4}$ $y=4^{-(-1)}=4$
0 $y =4^{0} = 1$ $y=4^{0}= 1$
1 $y =4^{1} = 4$ $y=4^{-1} = \frac{1}{4}$
2 $y =4^{2}=16$ $y=4^{-2}=\frac{1}{16}$
3 $y =4^{3} =64$ $y=4^{-3}= \frac{1}{64}$

Here is the graph of the two functions. Notice that the two functions are mirror images of each others if the mirror is placed vertically on the $y-$axis.

## Solve Real-World Problems Involving Exponential Decay

Exponential decay problems appear in several application problems. Some examples of these are half-life problems, and depreciation problems. Let’s solve an example of each of these problems.

Example 4 Half-Life

A radioactive substance has a half-life 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.

a) Draw the graph of the amount of radioactivity against time in weeks.

b) Find the formula that gives the radioactivity in terms of time.

c) Find the radioactivity left after three weeks

Solution

Let’s start by making a table of values and then draw the graph.

0 20
1 10
2 5
3 2.5
4 1.25
5 0.625

Exponential decay fits the general formula

$y=A \cdot b^x$

In this case

$y$ is the amount of radioactivity

$x$ is the time in weeks

$A=20$ is the starting amount

$b=\frac{1}{2}$ since the substance losses half its value each week

The formula for this problem is: $y = 20 \cdot \left (\frac{1} {2} \right)^x$ or $y = 20 \cdot 2^{-x}$.

Finally, to find out how much radioactivity is left after three weeks, we use $x=3$ in the formula we just found.

$y = 20 \cdot \left (\frac{1} {2} \right)^3 = \frac{20} {8} = 2.5$

Example 5 Depreciation

The cost of a new car is $32,000. It depreciates at a rate of 15% per year. This means that it looses 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 $t = 0$. Then we multiply the value of the car by 85% for each passing year. (Since the car looses 15% of its value, that means that it keeps 85% of its value). Remember that 85% means that we multiply by the decimal 0.85. 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 $y=A \cdot b^x$ In this case: $y$ is the value of the car $x$ is the time in years $A=32$ is the starting amount in thousands $b=0.85$ since we multiply the amount by this factor to get the value of the car next year The formula for this problem is $y=32 \cdot (0.85)^x$. Finally, to find the value of the car when it is four years old, we use $x=4$ in the formula we just found. $y=32 \cdot (0.85)^4=16.7$ thousand dollars or$16,704 if we don’t round.

## Review Questions

Graph the following exponential decay functions.

1. $y = \frac{1} {5}^x$
2. $y = 4 \cdot \left (\frac{2} {3} \right)^x$
3. $y = 3^{-x}$
4. $y = \frac{3} {4} \cdot 6^{-x}$

Solve the following application problems.

1. The cost of a new ATV (all-terrain 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 year old.
2. 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 $\frac{1}{4}$ of its size each day.
1. Draw the graph of the size of the bacteria population against time in days.
2. Find the formula that gives the size of the bacteria population in term of time.
3. Find the size of the bacteria population ten days after the drug was first taken.
4. Find the size of the bacteria population after 2 weeks (14 days)

1. Formula $y=7200 \cdot (0.82)^x$ At $x=10, y= \989.62$
1. Formula $y = 2,000,000 \cdot 4-x$ or $y = 2,000,000 \cdot (0.25)^x$
2. At $x=5, y=1953$ bacteria
3. At $x=10, y = 1.9$ ($\approx 2$ bacteria)
4. At $x = 14, y = 0.007$ (bacteria effectively gone)

Feb 22, 2012

Jul 24, 2014