Exponential Decay - Algebra 1

5.5.2 Exponential Decay

5.5.2 • Exponential Decay

Activity

Every year after a new car is purchased, it loses $\frac{1}{3}$ of its value. Another way to think about this is that the car retains $\frac{2}{3}$ of its value each year. Let’s say that a new car’s value is $18,000.

1. A buyer worries that the car will be worth nothing in three years. Do you agree? Be prepared to show your reasoning.

Your answers may vary, but here is an example:

Disagree. Sample explanation: If the car loses a third of its value each year, then after one year its value will be $\frac{2}{3}$ of $18,000 or $12,000. After two years its value will be $\frac{2}{3}$ of $12,000 or $8,000. And after three years its value will be $\frac{2}{3}$ of $8,000 or about $5,333. The buyer might have thought that “losing a third of its value each year” meant losing $6,000 each year, so after 3 years the car would be worthless (because $3 \cdot 6, 000 = 18, 000$ ). He may not realize that after one year, the value of the car is no longer $18,000.

2. Write an expression to show how to find the value of the car for each year listed in the table.

Year	Value of Car (Dollars)
0	18,000
1
2
3
6
$t$

Compare your answer:

Year	Value of Car (Dollars)
0	18,000
1	$18, 000 \cdot \frac{2}{3}$
2	$18, 000 \cdot \frac{2}{3} \cdot \frac{2}{3}$ or $18, 000 \cdot (\frac{2}{3})^{2}$
3	$18, 000 \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3}$ or $18, 000 \cdot (\frac{2}{3})^{3}$
6	$18, 000 \cdot (\frac{2}{3})^{6}$
$t$	$18, 000 \cdot (\frac{2}{3})^{t}$

3. Write an equation relating the value of the car in dollars, $v$ , to the number of years, $t$ .

Compare your answer:

$v = 18, 000 \cdot (\frac{2}{3})^{t}$

4. Use your equation to find $v$ when $t$ is 0. What does this value of $v$ mean in this situation?

Compare your answer:

$v = 18, 000$ . This is the value of the car when $t = 0$ , which is when the car is purchased.

5. A different car loses value at a different rate. The value of this different car in dollars, $d$ , after $t$ years can be represented by the equation $d = 10, 000 \cdot (\frac{4}{5})^{t}$ . Explain what the numbers 10,000 and $\frac{4}{5}$ mean in this situation.

Compare your answer:

10,000 is the initial value of the car in dollars, and the car retains $\frac{4}{5}$ of its value each year.

Are you ready for more?

Extending Your Thinking

Start with an equilateral triangle with area 1 square unit, divide it into 4 congruent pieces as in the figure, and remove the middle one. Then, repeat this process with each of the remaining pieces. Repeat this process over and over for the remaining pieces. The figure shows the first two steps of this construction.

Three blue triangles show the stages of forming a Sierpinski triangle: starting with a solid triangle, then removing the center, and finally removing the centers of the remaining triangles, creating a recursive pattern.

What fraction of the area is removed each time? How much area is removed after the $n^{th}$ step? Use a calculator to find out how much area remains in the triangle after 50 such steps have been taken.

Compare your answer:

Each step, $\frac{1}{4}$ of the remaining area is removed, since taking away $\frac{1}{4}$ from each smaller triangle results in taking $\frac{1}{4}$ of the total remaining area. That means that each step, the remaining area is multiplied by $\frac{3}{4}$ since $1 - \frac{1}{4} = \frac{3}{4}$ . So after $n$ steps, the remaining area is simply $(\frac{3}{4})^{n}$ . After 50 steps, this leaves an area of only about 0.000000566321656 square units, so there’s almost nothing left!

If you can imagine doing this process forever, the resulting shape is called the Sierpinski triangle.

$A large blue triangle filled with smaller blue triangles arranged in a repeating pattern, creating a fractal design known as the Sierpinski triangle. The inner triangles are inverted and evenly spaced throughout.$

Self Check

Which of the following could represent a population of bacteria,

b

, that decreases at a rate of

\frac18

every hour,

h

? Consider an initial amount of 100 bacteria.

$b=100 \cdot (\frac{1}{8})h$
$b=100 \cdot (\frac{7}{8})h$
$b=100 \cdot (\frac{1}{8})^h$
$b=100 \cdot (\frac{7}{8})^h$

Additional Resources

Analyzing Exponential Decay

At the beginning of April, a colony of ants had a population of 5,000.

The colony decreases by $\frac{1}{5}$ during April. Write an expression that represents the ant population at the end of April.
During May, the colony decreases again by $\frac{1}{5}$ of its size. Write an expression that represents the ant population at the end of May.
The colony continues to decrease by $\frac{1}{5}$ of its size each month. Write an expression that represents the ant population after 6 months.
Write an equation to represent the population of ants, $p$ , after $m$ months.

When you are analyzing exponential decay and are given the fraction or percent by which it decreases, you need to remember that one minus that fraction or percent is the growth rate. In the case of the ant colony, because it is decreasing by $\frac{1}{5}$ , that means that the growth factor is $1 - \frac{1}{5}$ or $\frac{4}{5}$ .

Part 1 asks for an expression to represent the ant population at the end of April. There are several options:

$5000 - (\frac{1}{5} \cdot 5000)$
$5000 \cdot \frac{4}{5}$
$4000$

Each line is the simplification of the line above it. Part 2 has a few more options.

Part 2 is asking you to find $\frac{4}{5}$ of April’s total. There are several expressions to represent that:

$5000 \cdot \frac{4}{5} \cdot \frac{4}{5}$
$4000 \cdot \frac{4}{5}$
$4000 - (\frac{1}{5} \cdot 4000)$
$3200$

Basically, you take the answer from April and multiply by $\frac{4}{5}$ again since the growth factor is $\frac{4}{5}$ . You could also look at it as taking the initial value times $\frac{4}{5}$ twice.

Part 3 asks you to extend the pattern to 6 months after the initial population was measured. After the $2^{nd}$ month, you multiplied the initial value by the growth factor twice, so for this situation you would need to multiply the initial value by the growth factor 6 times, or raise it to the sixth power. The possible answers would be $5000 \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5}$ or $5000 \cdot (\frac{4}{5})^{6}$ , which means that after 6 months the population would be about 1,311.

Part 4 wants the general equation to represent the situation. If you use the pattern from part 3, you can see that it can be written in the exponential form $y = a b^{x}$ , where $a$ is the initial value and $b$ is the growth factor. Using the variables suggested, the equation would be $p = 5000 (\frac{4}{5})^{m}$ .

Try it

Analyzing Exponential Decay

A doctor prescribes 125 milligrams of a therapeutic drug that decreases in effectiveness by $\frac{3}{10}$ each hour.

a. Write an expression to represent how much is left after 1 hour.

b. Write an expression to represent how much is left after 3 hours.

c. Write an equation that represents the amount of drug, $d$ , left after $h$ hours.

Compare your answers:

Here is how to analyze the exponential decay situation:

In part a, you need to multiply the initial amount by $\frac{7}{10}$ , or $125 \cdot \frac{7}{10}$ , or 87.5 mg.

In part b, since it’s 3 hours later, it would be $125 \cdot \frac{7}{10} \cdot \frac{7}{10} \cdot \frac{7}{10}$ , or $125 \cdot (\frac{7}{10})^{3}$ , or 42.875 mg.

The equation that represents the situation for part c is $d = 125 \cdot (\frac{7}{10})^{h}$ .