6.025 Exponential growth and decay

Exponential growth and decay

An exponential function is the appropriate model to use when a quantity is increasing or decreasing at a rate that depends on the quantity present.

For example, in the final rounds of a sports competition, the number of competing teams is halved at every stage. Thus, if $16$16 teams reached the first semifinal, there would be $8$8 in the second semifinal, and so on. The number of teams playing drops from $16$16 to $8$8 and then to $4$4 and finally, $2$2 and we see that the reduction in the number of teams playing at each stage depends on the number in the previous round.

This is an example of exponential decay. The rate of decrease gets progressively smaller. Many processes show the opposite pattern and exhibit exponential growth. In this, the rate of increase increases progressively.

One of the most common applications is where there is a percentage growth or a percentage decay. For example the amount in a savings account or the remaining balance on a loan respectively. Let's look at the general formula for these scenarios.

Exploration

A particular sports convertible costs $\$48000$$48000. It loses $18%$18% of its value per year. We want to find its value for the next $4$4 years. Let's use a table to do so.

Year	Amount Lost	Value ($\$$$)	Expression for Value
$0$0	n/a	$48000$48000	$48000$48000
$1$1	$0.18\times48000=8640$0.18×48000=8640	$48000-8640=39360$48000−8640=39360	$48000-48000\times0.18=48000\left(1-0.18\right)$48000−48000×0.18=48000(1−0.18)
$2$2	$0.18\times39360=7084.80$0.18×39360=7084.80	$39360-7084.80=32275.20$39360−7084.80=32275.20	$48000\left(1-0.18\right)^2$48000(1−0.18)2
$3$3	$0.18\times32275.20=5809.54$0.18×32275.20=5809.54	$32275.20-5809.54=26465.66$32275.20−5809.54=26465.66	$48000\left(1-0.18\right)^3$48000(1−0.18)3
$4$4	$0.18\times26465.66=4763.82$0.18×26465.66=4763.82	$26465.66-4763.82=21701.84$26465.66−4763.82=21701.84	$48000\left(1-0.18\right)^4$48000(1−0.18)4

Rather than continuing to work through each year in this way, we could approach the problem differently by looking at the column Expression for value.

Notice that multiplying by $18%$18% and then subtracting that from the current value is the same as multiplying by $\left(1-0.18\right)=0.82$(1−0.18)=0.82. Since we are doing this repeatedly, we can make a formula.

In fact, for exponential growth and decay by a percentage increase or decrease, we can use the formula below:

Worked examples

Example 1

Justine bought an oil painting for $\$1600$$1600 whose value is given by the formula $A=1600\times0.94^t$A=1600×0.94t, and $t$t is the number of years passed.

(a) What's the annual depreciation rate?

Think: What's $r$r in this case?

Do:

$1600\times0.94^t=1600\left(1-0.06\right)^t$1600×0.94t=1600(1−0.06)t

so the annual depreciation rate is $0.06=6%$0.06=6%.

(b) How much would the painting be worth in 10 years, round to the nearest dollar?

Think: What parameters do we have, which one are we looking for. It can be helpful to list values before jumping into the formula. For the equation, $A=1600\times0.94^t$A=1600×0.94t, we are given $t=10$t=10 and are looking for $A$A.

Do:

$A$A	$=$=	$1600\times0.94^t$1600×0.94t
	$=$=	$1600\times0.94^{10}$1600×0.9410
	$=$=	$861.7841826$861.7841826

 

So the painting would be worth $\$862$$862 in $10$10 years.

example 2

A certain radioactive isotope decays in such a way that after $175$175 years only half of the original quantity of the isotope remains. Suppose $10$10 kg of the substance existed initially. Create an equation for the amount of the isotope left after $t$t years.

Think: When given a worded problem, it helps to list what we were given and what we are looking for.

$A$A: This will be an unknown based on the number of years, $t$t

$P$P: We initially have $10$10 kg, so $P=10$P=10

$r$r: This is the rate of decay, which we can say is $50%$50% if we are looking at the half-life

$n$n: This will be filled in based on the time, we need to remember that the rate is in groups of $175$175 years, so $n$n must be in groups of 175 years

Do:

After $175$175 years, only $5$5 kg will be left and then after a further $175$175 years, only $2.5$2.5 kg will be left, and so on. After $n$n groups of $175$175 years, we have that:

$A=10\times\left(\frac{1}{2}\right)^n$A=10×(12​)n

The number of years that have elapsed since the beginning of this experiment must be $t=175n$t=175n. So, we can put $n=\frac{t}{175}$n=t175​ into the formula so that we no longer have to convert the time elapsed into groups of $175$175 years but can use just years instead.

Thus, we arrive at a formula for the amount remaining after $t$t years:

$A=10\times\left(\frac{1}{2}\right)^{\frac{t}{175}}$A=10×(12​)t175​

Reflect: How could we use this equation? How much would be left after $1000$1000 years?

Practice questions

Question 1

question 2