Math Core

Banks advertise interest as a percent, towns report population growth as a percent, and a car loses a percent of its value each year. Whenever a quantity changes by the same percent in each time period, it follows an exponential model. This lesson turns percent rates into growth factors and puts the models to work.

From a percent to a growth factor

Suppose a town of $2000$ people grows by $5\%$ per year. After one year it has its original population plus $5\%$ more:

2000 + 0.05(2000) = 2000(1 + 0.05) = 2000(1.05) = 2100.

Increasing by $5\%$ is the same as multiplying by $1.05$ . The next year the population is multiplied by $1.05$ again, so after $t$ years it is $2000(1.05)^t$ .

Decreasing works the same way. Losing $15\%$ leaves $100\% - 15\% = 85\%$ , so a $15\%$ decrease means multiplying by $0.85$ .

Exponential growth and decay models

A quantity with initial amount $a$ that changes at a rate $r$ (written as a decimal) per time period:

\text{growth: } y = a(1 + r)^t \qquad\qquad \text{decay: } y = a(1 - r)^t

Here $t$ is the number of time periods. The growth factor $1 + r$ is greater than $1$ ; the decay factor $1 - r$ is between $0$ and $1$ .

Common mistake

Convert the percent to a decimal before building the factor. A $4\%$ growth rate gives a factor of $1.04$ , not $1.4$ (that would be $40\%$ ) and not $4$ .

For decay, don't use the rate itself as the factor. A $20\%$ decrease multiplies by $0.80$ , not $0.20$ . Using $0.20$ would mean losing $80\%$ each period.

Reading a model

You can also work backward from an equation to the rate.

Worked example: Identify the rate

For each model, tell whether it shows growth or decay, and give the percent rate.

$y = 450(1.12)^t$
$y = 3000(0.94)^t$

Solutions.

The factor $1.12 = 1 + 0.12$ is greater than $1$ : growth at $12\%$ per period. The initial amount is $450$ .
The factor $0.94 = 1 - 0.06$ is less than $1$ : decay at $6\%$ per period. The initial amount is $3000$ .

Applying growth and decay

Worked example: Depreciation

A new car costs $18,000 and loses $15\%$ of its value each year. What is it worth after $3$ years?

The decay factor is $1 - 0.15 = 0.85$ :

y = 18000(0.85)^3 = 18000(0.614125) = 11054.25.

The car is worth $11,054.25 after $3$ years. Notice it lost about $6,946, which is less than $3 \times 15\% \times 18000 = 8100$ . Each year's $15\%$ is taken from a smaller value.

Doubling time and half-life

Sometimes a rate is described by how long it takes to double or to halve. Then the factor is $2$ or $\tfrac{1}{2}$ , and the exponent counts how many doubling (or halving) periods have passed.

y = a \cdot 2^{t/d} \qquad\qquad y = a\left(\tfrac{1}{2}\right)^{t/h}

where $d$ is the doubling time and $h$ is the half-life, in the same units as $t$ .

Worked example: Half-life

A medicine has a half-life of $6$ hours. A patient takes an $80$ mg dose. How much remains after $24$ hours?

In $24$ hours there are $\dfrac{24}{6} = 4$ half-lives. Halve $80$ four times:

80\left(\tfrac{1}{2}\right)^4 = 80 \cdot \tfrac{1}{16} = 5 \text{ mg}.

Step by step: $80 \to 40 \to 20 \to 10 \to 5$ .

Compound interest

When a bank pays interest, the interest is added to your balance, and next time you earn interest on the interest too. That's called compounding, and it's exponential growth.

If the interest is compounded once a year, the model is just $A = P(1 + r)^t$ , where $P$ is the principal (starting amount). Banks often compound more often, splitting the yearly rate into $n$ equal pieces:

A = P\left(1 + \frac{r}{n}\right)^{nt}

For monthly compounding $n = 12$ , so each month the balance is multiplied by $1 + \tfrac{r}{12}$ , and over $t$ years that happens $12t$ times.

Worked example: Compounding monthly

You deposit $2,000 in an account that pays $6\%$ annual interest compounded monthly. How much is in the account after $2$ years?

Here $P = 2000$ , $r = 0.06$ , $n = 12$ and $t = 2$ . The monthly factor is $1 + \tfrac{0.06}{12} = 1.005$ , applied $12 \cdot 2 = 24$ times:

A = 2000(1.005)^{24} \approx 2000(1.127160) \approx 2254.32.

The balance is about $2,254.32. Compounding once a year instead would give $2000(1.06)^2 = 2247.20$ , a little less.

Tip

Use your calculator's power key ( $\wedge$ or $y^x$ ) and round only at the very end. Rounding the growth factor early, like writing $(1.005)^{24} \approx 1.13$ , can throw a money answer off by several dollars.

Practice

Practice 1

A town's population grows by $5\%$ each year. By what number is the population multiplied each year?