Interpreting Negative Exponents in Exponential Decay - Algebra 1

5.6.3 Interpreting Negative Exponents in Exponential Decay

5.6.3 • Interpreting Negative Exponents in Exponential Decay

Activity

A person took some medicine but does not remember how much. Concerned that she took too much, she has a blood test every hour for several hours.

$t$ , time (hours)	$m$ , medicine (mg)
–3
–1
0	100
1	50
2	25

Answer the following questions.

1. Time $t$ is measured in hours since the blood test, and amount of medicine in her body, $m$ , is measured in milligrams. What is the decay factor? That is, what is $b$ in an equation of the form $m = a \cdot b^{t}$ ?

Compare your answer:

$b = \frac{1}{2}$

2. What is $a$ ?

Compare your answer:

$a = 100$

3. Find the amounts of medicine in the patient’s body when $t$ is –1.

Compare your answer:

When $t$ is –1, $m$ is 200.

4. Find the amounts of medicine in the patient’s body when $t$ is –3.

Compare your answer:

When $t$ is –3, $m$ is 800.

5. What does $t = 0$ mean in this context?

Compare your answer:

$t = 0$ means the time the doctor did the first blood test.

6. What does $t = - 3$ mean in this context?

Compare your answer:

$t = - 3$ means 3 hours before the doctor did the first test.

7. The medicine was taken when $t$ is –5. Assuming the person did not have any of the medication in her body beforehand, how much medicine did the patient take?

Compare your answer:

3,200 mg. Each hour, this is cut in half. After being halved 5 times, it leaves 100 mg, the amount of medicine in the person’s body when $t = 0$ .

8. Create a graph by plotting the points whose coordinates are in the table from question 1 above. Make sure to draw and label tick marks on the axes.

A graph with a vertical axis labeled m and a horizontal axis labeled t, both originating from point O at the bottom left corner; the axis lines are bolded, but there are no data or additional markings.

Compare your answer:

A scatter plot with points showing high values on the left (negative t) and low values on the right (positive t). The vertical axis is labeled m and the horizontal axis is labeled t with the origin marked as O. Points plotted at (-3, 800), (-2, 400), (-1, 200), (0, 100), and assympotal to the x-axis towards x=3.

9. Based on your graph, when do you think the patient will have: 500 mg of medicine remaining in her body

Compare your answer:

Between 2 and 3 hours before the patient first had her blood tested

10. Based on your graph, when do you think the patient will have: No medicine remaining in her body

Compare your answer: Your answer may vary, but here are some samples:

Sometime after 10 hours. The amount of medicine falls below 0.1 mg after about 10 hours and might be too small to be detectable after that point.
Eventually, all of the medicine will be out of the patient’s body (but the mathematical model for the amount of medicine does not reach 0).

Are you ready for more?

Extending Your Thinking

Without evaluating them, describe each of the following quantities as close to 0, close to 1, or much larger than 1.

1. $\frac{1}{1 - 2^{- 10}}$

close to 1

Without evaluating them, describe each of the following quantities as close to 0, close to 1, or much larger than 1.

2. $\frac{2^{10}}{2^{10} + 1}$

close to 1

Without evaluating them, describe each of the following quantities as close to 0, close to 1, or much larger than 1.

3. $\frac{2^{- 10}}{2^{10} + 1}$

close to 0

Without evaluating them, describe each of the following quantities as close to 0, close to 1, or much larger than 1.

4. $\frac{1 - 2^{- 10}}{2^{10}}$

close to 0

Without evaluating them, describe each of the following quantities as close to 0, close to 1, or much larger than 1.

5. $\frac{1 + 2^{10}}{2^{- 10}}$

much larger than 1

Self Check

A census of hemlock trees is taken in a national park. The rangers have found that only $\frac45$ of the population

$y$ , years since census	$h$ , number of hemlocks
–2
0	64,000
1	51,200
3	32,768

100,000
80,000
125,000
156,250

Additional Resources

Interpreting Negative Exponents in Exponential Decay

Let’s look at a problem involving an exponential decay equation and negative exponents.

A scientist records the population of a bacterial population in a petri dish. The data are shown in the table.

$d$ , days since first test	$b$ , number of bacteria
–3
–1
0	40,000
1	20,000
3	5,000

The number of days $d$ since the first test corresponds with the number of bacteria, $b$ , estimated to be alive in the petri dish.

1. What is the decay factor? That is, what is $f$ in an equation of the form $b = n \cdot f^{d}$ ? What is the starting population, $n$ , on the first day of testing?

Since the number of bacteria found on day 0 is 40,000, then $n = 40, 000$ . Since after each day, the number of bacteria is cut in half, then the decay factor, $f$ , is $\frac{1}{2}$ .

The equation to represent this situation is $b = 40, 000 \cdot {(\frac{1}{2})}^{d}$ .

2. What do $d = - 1$ and $d = - 3$ mean in this context?

Since the day the first test is taken is defined as $d = 0$ , then days with a negative value occured before the first test. $d = - 1$ means 1 day before the scientist did the first test. $d = - 3$ means 3 days before the scientist did the first test.

3. How many bacteria were present 1 day before the first test?

Substitute –1 for $d$ in the equation $b = 40, 000 \cdot {(\frac{1}{2})}^{d}$ .

$b = 40, 000 \cdot {(\frac{1}{2})}^{d}$

$b = 40, 000 \cdot {(\frac{1}{2})}^{- 1}$

$b = 40, 000 \cdot 2$

$b = 80, 000$

There were 80,000 bacteria in the petri dish 1 day before the test.

4. Assume the peak of the bacterial population was 3 days before the first test. How many bacteria were present at the peak of the population?

Substitute –3 for $d$ in the equation $b = 40, 000 \cdot {(\frac{1}{2})}^{d}$ .

$b = 40, 000 \cdot {(\frac{1}{2})}^{d}$

$b = 40, 000 \cdot {(\frac{1}{2})}^{- 3}$

$b = 40, 000 \cdot 8$

$b = 320, 000$

There were 320,000 bacteria in the petri dish 3 days before the test at the peak of the population.

Try it

Interpreting Negative Exponents in Exponential Decay

An investor has tracked the value of a particular investment since 2016 that has depreciated over time. The data are shown in the table.

$y$ , years since 2016	$v$ , value of investment ($)
–2
–1
0	36,000
1	12,000
2	4,000

What is the equation that represents this situation?
What do $y = - 1$ and $y = - 2$ mean in this context?
If the value of the investment was at its highest in 2014, what was the maximum value of the investment?

Compare your answers:

Here is how to solve this exponential growth problem using negative exponents:

The equation that represents this situation is $v = 36, 000 \cdot {(\frac{1}{3})}^{y}$ . The value of the investment in year 0 is $36,000, and the decay factor is $\frac{1}{3}$ .
$y = - 1$ means 1 year before 2016, which is 2015. $y = - 2$ means 2 years before 2016, which is 2014.
The maximum value of the investment was $324,000. $v = 36, 000 \cdot {(\frac{1}{3})}^{y}$ $v = 36, 000 \cdot {(\frac{1}{3})}^{- 2}$ $v = 36, 000 \cdot 9$ $v = 324, 000$