8.6: Exponential Decay Functions
In the last lesson, we learned how to solve expressions that modeled exponential growth. In this lesson, we will be learning about exponential decay functions.
General Form of an Exponential Function: \begin{align*}y=a (b)^x\end{align*}
\begin{align*}b = growth \ factor\end{align*}
In exponential decay situations, the growth factor must be a fraction between zero and one.
\begin{align*}0<b<1\end{align*}
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. Nadia loves candy, so she ate half the bag on the first day. Her mother told her that if she continues to eat at that rate, it will be gone the next day and she will not have any more 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?
Solution: Make a table of values for this problem.
Day | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|---|
# 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 has only 1.25 candies left in her bag.
Write an equation for this exponential function. Nadia began with 160 pieces of candy. In order to get the amount of candy left at the end of each day, we keep multiplying by \begin{align*}\frac{1}{2}\end{align*}
\begin{align*}y=160 \cdot \frac{1}{2}^x\end{align*}
Graphing Exponential Decay Functions
Example: Graph the exponential function \begin{align*}y=5 \cdot \left(\frac{1}{2}\right)^x\end{align*}
Solution: Start by making a table of values. Remember when you have a number to the negative power, you are simply taking the reciprocal of that number and taking it to the positive power. Example: \begin{align*}\left(\frac{1}{2}\right)^{-2} = \left(\frac{2}{1}\right)^2 = 2^2\end{align*}
\begin{align*}x\end{align*} |
\begin{align*}y=5 \cdot \left(\frac{1}{2}\right)^x\end{align*} |
---|---|
–3 |
\begin{align*}y =5 \left(\frac{1}{2}\right)^{-3}=40\end{align*} |
–2 |
\begin{align*}y =5 \left(\frac{1}{2}\right)^{-2}=20\end{align*} |
–1 |
\begin{align*}y =5 \left(\frac{1}{2}\right)^{-1}=10\end{align*} |
0 |
\begin{align*}y =5 \left(\frac{1}{2}\right)^0=5\end{align*} |
1 |
\begin{align*}y =5 \left(\frac{1}{2}\right)^1=\frac{5}{2}\end{align*} |
2 |
\begin{align*}y =5 \left(\frac{1}{2}\right)^2=\frac{5}{4}\end{align*} |
Now graph the function.
Using the Property of Negative Exponents, the equation can also be written as \begin{align*}5 \cdot 2^{-x}\end{align*}
Comparing Graphs of Exponential Decay Functions
Exponential growth and decay graphs look like opposites and can sometimes be mirror images.
Example: Graph the functions \begin{align*}y=4^x\end{align*}
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 “reverse images” of each other in the sense that the values for the two functions are reciprocals.
\begin{align*}x\end{align*} |
\begin{align*}y=4^x\end{align*} |
\begin{align*}y=4^{-\chi}\end{align*} |
---|---|---|
–3 |
\begin{align*}y=4^{-3} = \frac{1}{64}\end{align*} |
\begin{align*}y=4^{-(-3)} = 64\end{align*} |
–2 |
\begin{align*}y=4^{-2} = \frac{1}{16}\end{align*} |
\begin{align*}y=4^{-(-2)} = 16\end{align*} |
–1 | \begin{align*}y=4^{-1} = \frac{1}{4}\end{align*} | \begin{align*}y=4^{-(-1)} = 4\end{align*} |
0 | \begin{align*}y=4^0 = 1\end{align*} | \begin{align*}y=4^{-(0)} = 1\end{align*} |
1 | \begin{align*}y=4^1 = 4\end{align*} | \begin{align*}y=4^{-(1)} =\frac{1}{4}\end{align*} |
2 | \begin{align*}y=4^2 = 16\end{align*} | \begin{align*}y=4^{-(2)} =\frac{1}{16}\end{align*} |
3 | \begin{align*}y=4^3 = 64\end{align*} | \begin{align*}y=4^{-(3)} = \frac{1}{64}\end{align*} |
Here is the graph of the two functions. Notice that these two functions are mirror images if the mirror is placed vertically on the \begin{align*}y-\end{align*}axis.
Solving Real-World Problems Involving Exponential Decay Functions
Example: 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 the 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.
Solution: 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.
\begin{align*}y=32,000(0.85)^4=16,704.\end{align*}
Practice Set
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: Exponential Decay Functions (10:51)
- Define exponential decay.
- What is true about “b” in an exponential decay function?
- Suppose \begin{align*}f(x)=a(b)^x\end{align*}. What is \begin{align*}f(0)\end{align*}? What does this mean in terms of the \begin{align*}y-\end{align*}intercept of an exponential function?
Graph the following exponential decay functions.
- \begin{align*}y=\frac{1}{5}^x\end{align*}
- \begin{align*}y=4 \cdot \left(\frac{2}{3}\right)^x\end{align*}
- \begin{align*}y=3^{-x}\end{align*}
- \begin{align*}y=\frac{3}{4} \cdot 6^{-x}\end{align*}
Solve the following application problems.
- 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 years old.
- Michigan’s population is declining at a rate of 0.5% per year. In 2004, the state had a population of 10,112,620.
- Write a function to express this situation.
- If this rate continues, what will the population be in 2012?
- When will the population of Michigan reach 9,900,000?
- What was the population in the year 2000, according to this model?
- A certain radioactive substance has a half-life of 27 years. An organism contains 35 grams of this substance on day zero.
- Draw the graph of the amount remaining. Use these values for \begin{align*}x: x=0, 27, 54, 81, 108, 135.\end{align*}
- Find the function that describes the amount of this substance remaining after \begin{align*}x\end{align*} days.
- Find the amount of radioactive substance after 92 days.
- The percentage of light visible at \begin{align*}d\end{align*} meters is given by the function \begin{align*}V(d)=0.70^d\end{align*}.
- What is the growth factor?
- What is the initial value?
- Find the percentage of light visible at 65 meters.
- 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 two weeks (14 days).
Mixed Review
- The population of Kindly, USA is increasing at a rate of 2.14% each year. The population in the year 2010 is 14,578.
- Write an equation to model this situation.
- What would the population of Kindly be in the year 2015?
- When will the population be 45,000?
- The volume of a sphere is given by the formula \begin{align*}v=\frac{4}{3} \pi r^3\end{align*}. Find the volume of a sphere with a diameter of 11 inches.
- Simplify \begin{align*}\frac{6x^2}{14y^3} \cdot \frac{7y}{x^8} \cdot x^0 y\end{align*}.
- Simplify \begin{align*}3(x^2 y^3 x)^2\end{align*}.
- Rewrite in standard form: \begin{align*}y-16+x=-4x+6y+1\end{align*}.