A town of $12000$ people grows by $3\%$ a year. Write a model and find the population after $5$ years.

$P = 12000(1.03)^n$; about $13911$ people after $5$ years.

A sample of $80$ mg has a half-life of $5$ days. How much is left after $20$ days, and when are $2.5$ mg left?

$5$ mg after $20$ days; $2.5$ mg after $25$ days.

$5000$ dollars is invested at $4.8\%$ a year, compounded monthly, for $6$ years. Find the final amount and the interest earned, and compare with compounding once a year.

About $6664.96$ dollars, with $1664.96$ dollars of interest: $40.69$ dollars more than yearly compounding gives.

Grade 11 · MCR3U · Functions

MCR3U: Exponential Growth and Decay

Exponential growth and decay describe anything that changes by the same percent in each equal stretch of time: a population, a bank balance, a car's value, a radioactive sample. One model handles all of them, and the work is reading the question into it.

How It Works

The model. $A = A_0(1 + r)^n$ , where $A_0$ is the starting amount, $r$ is the rate per period as a decimal and $n$ is the number of periods. Growth has $r > 0$ . Decay has $r < 0$ , so a $15\%$ loss each year gives a factor of $1 - 0.15 = 0.85$ .

Read the growth factor. In $A = 500(1.04)^n$ the amount grows $4\%$ per period, and in $A = 500(0.9)^n$ it shrinks $10\%$ . A factor above $1$ is growth and a factor between $0$ and $1$ is decay.

Doubling time and half-life. When an amount doubles every $d$ units of time, $A = A_0(2)^{\frac{t}{d}}$ . When it halves every $h$ units, $A = A_0\left(\frac{1}{2}\right)^{\frac{t}{h}}$ . The exponent counts how many doublings or half-lives have passed.

Compound interest is the same model, with the periods set by the bank: $A = P(1 + i)^n$ . Here $i$ is the annual rate divided by the number of compounding periods per year, and $n$ is the total number of periods. $5\%$ a year compounded monthly for $3$ years gives $i = \frac{0.05}{12}$ and $n = 36$ .

Solving for time. When the unknown is in the exponent, write both sides as powers of the same base if you can: $2^{\frac{t}{3}} = 16 = 2^4$ , so $\frac{t}{3} = 4$ and $t = 12$ . When the numbers do not allow that, use a graph or trial and improvement.

Answering. Round only at the end, and give units. Money goes to the nearest cent, and a count of people or cells to a whole number.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
A town of $12000$ people grows by $3\%$ a year. Write a model and find the population after $5$ years.
Show solution
1. The factor is $1 + 0.03 = 1.03$ , so $P = 12000(1.03)^n$ .
2. $P = 12000(1.03)^5 \approx 13911.3$ .
Answer: $P = 12000(1.03)^n$ ; about $13911$ people after $5$ years.
Easy
A culture of $250$ bacteria doubles every $4$ hours. How many bacteria are there after $1$ day?
Show solution
1. $1$ day is $24$ hours, which is $\frac{24}{4} = 6$ doublings.
2. $A = 250(2)^6 = 250(64)$ .
Answer: $16000$ bacteria.
Medium
A sample of $80$ mg has a half-life of $5$ days. How much is left after $20$ days, and when are $2.5$ mg left?
Show solution
1. $A = 80\left(\frac{1}{2}\right)^{\frac{t}{5}}$ . At $t = 20$ : $80\left(\frac{1}{2}\right)^4 = \frac{80}{16} = 5$ .
2. $80\left(\frac{1}{2}\right)^{\frac{t}{5}} = 2.5$ gives $\left(\frac{1}{2}\right)^{\frac{t}{5}} = \frac{1}{32} = \left(\frac{1}{2}\right)^5$ .
3. So $\frac{t}{5} = 5$ and $t = 25$ .
Answer: $5$ mg after $20$ days; $2.5$ mg after $25$ days.
Medium
A car bought for $32000$ dollars loses $15\%$ of its value each year. What is it worth after $6$ years?
Show solution
1. The factor is $1 - 0.15 = 0.85$ , so $V = 32000(0.85)^n$ .
2. $V = 32000(0.85)^6 \approx 32000(0.377150) \approx 12068.78$ .
Answer: About $12068.78$ dollars.
Test level
$5000$ dollars is invested at $4.8\%$ a year, compounded monthly, for $6$ years. Find the final amount and the interest earned, and compare with compounding once a year.
Show solution
1. Monthly: $i = \frac{0.048}{12} = 0.004$ and $n = 6 \times 12 = 72$ .
2. $A = 5000(1.004)^{72} \approx 6664.96$ , so the interest is $6664.96 - 5000 = 1664.96$ .
3. Yearly: $A = 5000(1.048)^6 \approx 6624.27$ .
4. Monthly compounding earns $6664.96 - 6624.27 = 40.69$ more.
Answer: About $6664.96$ dollars, with $1664.96$ dollars of interest: $40.69$ dollars more than yearly compounding gives.

Learn It on Martian Lab

Martian Lab teaches this in the Exponential Applications lesson of its MCR3U Exponential Functions unit. Watch it, then take the Exponential Functions practice tests to see if it stuck.

Watch the Exponential Applications lesson free

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Free Exponential Functions practice test, with solutions

More Exponential Functions topics: Exponent Laws and Rational Exponents · Graphing Exponential Functions

More MCR3U topics: Inverse Functions · Transformations of Functions · Sine Law and the Ambiguous Case · Domain, Range and Function Notation · Simplifying Rational Expressions · Operations with Rational Expressions · Max/Min and Vertex Form · Linear-Quadratic Systems · Quadratic Word Problems · Simplifying Radicals · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Transformations of Sine and Cosine · Sinusoidal Word Problems · Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest · Annuities: Future and Present Value

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