Exponential Growth and Decay, SAT math lesson

A savings account that earns interest, a culture of bacteria and the value of a used car all change by the same percent over each equal stretch of time, not by the same amount. That repeated multiplying is what an exponential model describes. Most SAT questions on this topic come down to reading three things correctly in a formula, where it starts, what it multiplies by, and over what length of time.

See also Exponential Functions with Different Bases in Foundations for Calculus and Exponential Growth and Decay Models in Precalculus.

Key terms

Exponential model

Definition An exponential model is a function of the form

f(t) = a \cdot b^t,

where $a \ne 0$ and the base $b$ is a positive real number other than $1$ . The base must be positive so that $b^t$ is a real number for every real $t$ , including fractional exponents such as $t = \frac{1}{2}$ . In applications $a > 0$ and $t$ counts units of time from a starting moment. A model describes a real situation only under its stated assumptions, such as a constant percent change and the range of times it was built for.

In plain words Over each unit of time the quantity is multiplied by the same number $b$ . A linear model adds the same amount each step, an exponential model multiplies by the same factor.

Example A town has $f(t) = 5000 \cdot (1.02)^t$ residents $t$ years after 2020. Each year the population is $1.02$ times the year before, so in 2022 it is $5000 \cdot 1.02^2 = 5202$ . Using the same formula for 300 years would mean assuming the growth rate never changes, which no one should take for granted.

Initial value

Definition The initial value of the model $f(t) = a \cdot b^t$ is $f(0) = a \cdot b^0 = a$ . It is the value of the quantity at $t = 0$ .

In plain words The number in front is where the quantity starts, measured at the moment the clock reads zero.

Example A car bought for $24{,}000$ dollars loses value according to $V(t) = 24{,}000 \cdot (0.88)^t$ after $t$ years. The initial value $24{,}000$ is the purchase price. After two years the value is $24{,}000 \cdot 0.88^2 = 18{,}585.60$ dollars.

Growth factor and decay factor

Definition In $f(t) = a \cdot b^t$ with $a > 0$ , the base $b$ is the factor per unit of time, because

\frac{f(t + 1)}{f(t)} = b \quad \text{for every } t.

If $b > 1$ the quantity grows and $b$ is called the growth factor. If $0 < b < 1$ the quantity decays and $b$ is called the decay factor. Data with equally spaced inputs and positive outputs fit an exponential model exactly when the ratio of consecutive outputs is the same constant every time, and that constant is not $1$ .

In plain words Divide a value by the one before it. If you always get the same number, that number is the factor.

Example A table of a quantity at whole hours is shown below.

$t$	$0$	$1$	$2$	$3$
$f(t)$	$400$	$500$	$625$	$781.25$

The ratios are $\frac{500}{400} = \frac{625}{500} = \frac{781.25}{625} = 1.25$ , while the differences $100$ , $125$ and $156.25$ are not equal. So the data are exponential with growth factor $1.25$ , and $f(t) = 400 \cdot (1.25)^t$ .

Percent rate

Definition If a quantity increases by $r$ percent over each time interval, its factor for that interval is

b = 1 + \frac{r}{100}.

If it decreases by $r$ percent over each interval, with $0 < r < 100$ , its factor is

b = 1 - \frac{r}{100}.

A percent rate always belongs to a specific time interval, such as per year or per month.

In plain words Growing by $6\%$ means keeping all of it and adding $6\%$ , so multiply by $1.06$ . Shrinking by $15\%$ means keeping $85\%$ , so multiply by $0.85$ .

Example A lake holds $1200$ fish, and the number falls by $15\%$ each year. The model is $F(t) = 1200 \cdot (0.85)^t$ , and after two years $F(2) = 1200 \cdot 0.7225 = 867$ fish. The decay factor is $0.85$ and the rate is $15\%$ per year. A common trap is to read $0.85$ as an $85\%$ decrease.

Time interval

Definition If a quantity is multiplied by $b$ once every $k$ units of time, with $k > 0$ , then after $t$ units it is

f(t) = a \cdot b^{t/k}.

The exponent $\frac{t}{k}$ counts how many full or partial intervals of length $k$ have passed.

In plain words Doubling every 3 hours does not mean doubling every hour. Divide the elapsed time by the length of one interval to get the number of doublings.

Example A culture of $500$ cells doubles every $3$ hours, so $N(t) = 500 \cdot 2^{t/3}$ with $t$ in hours. After $12$ hours there have been $\frac{12}{3} = 4$ doublings, and

N(12) = 500 \cdot 2^4 = 8000 \text{ cells.}

Equivalent exponential form

Definition For $b > 0$ and real numbers $m$ and $n$ , the power rule $b^{mn} = (b^m)^n$ lets one model be written in several equivalent forms. In particular, for $k > 0$

b^{t/k} = \left(b^{1/k}\right)^t,

so a factor $b$ per $k$ units is the same as a factor $b^{1/k}$ per unit. Each form shows the factor for a different time interval.

In plain words The same growth can be described per hour, per day or per year. Powers and roots convert one description to another.

Example A balance grows by $1\%$ per month. After $t$ years it has been multiplied by

1.01^{12t} = \left(1.01^{12}\right)^t \approx 1.1268^t.

So the yearly growth is about $12.68\%$ , not $12\%$ , because each month’s growth is applied to a larger amount.

Time-unit conversion

Definition To rewrite a model $f(t) = a \cdot b^t$ , with $t$ in one time unit, in terms of a new variable $s$ in another unit, write $t$ in terms of $s$ and substitute. If one old unit equals $n$ new units, then $t = \frac{s}{n}$ and

a \cdot b^{s/n} = a \cdot \left(b^{1/n}\right)^s.

In plain words Replace the old time variable by its expression in the new unit, then simplify the base with a power or a root.

Example An investment is worth $P(t) = 800 \cdot (1.21)^t$ dollars after $t$ years. With $h$ the number of half years, $t = \frac{h}{2}$ , so

P = 800 \cdot (1.21)^{h/2} = 800 \cdot (1.1)^h.

The value grows by $10\%$ per half year. Both forms agree. After $3$ years, or $6$ half years, each gives about $1417.25$ dollars.

Exponential equations with a common base

Definition An exponential equation has the unknown in an exponent. If $b > 0$ and $b \ne 1$ , then for all real $m$ and $n$

b^m = b^n \quad \text{exactly when} \quad m = n.

So if both sides can be written as powers of the same base, set the exponents equal. The condition $b \ne 1$ matters, since $1^m = 1^n$ for every $m$ and $n$ .

In plain words Rewrite both sides with the same base. Then the exponents must match, and you are left with an ordinary equation.

Example Solve $2^{x + 1} = 16$ . Since $16 = 2^4$ , the equation is $2^{x + 1} = 2^4$ , so $x + 1 = 4$ and $x = 3$ . The same idea recovers a missing time. A $3000$ milligram sample has a half-life of $5$ hours, so the amount after $t$ hours is $3000 \cdot \left(\frac{1}{2}\right)^{t/5}$ . To find when $375$ milligrams remain, divide by $3000$ .

\left(\frac{1}{2}\right)^{t/5} = \frac{375}{3000} = \frac{1}{8} = \left(\frac{1}{2}\right)^3.

So $\frac{t}{5} = 3$ and $t = 15$ hours. In the Desmos graphing calculator you could also graph y=3000(1/2)^(x/5) and y=375 and click their intersection point, which is at $(15, 375)$ .

Check yourself

The value of a car $t$ years after purchase is $V(t) = 24{,}000 \cdot (0.88)^t$ dollars. Which statement best describes the number $0.88$ ? (A) The value decreases by $88\%$ each year. (B) The value decreases by $12\%$ each year. (C) The value decreases by $0.88$ dollars each year. (D) The value decreases by $12\%$ over the whole period.

Show the answer

The correct choice is (B). Each year the value is multiplied by $0.88 = 1 - 0.12$ , so it keeps $88\%$ and loses $12\%$ of the previous year’s value. Choice (A) confuses the part kept with the part lost. Choice (C) describes a linear model, and (D) ignores that the factor applies once per year.

What value of $x$ satisfies $5^{2x - 1} = 125$ ?

Show the answer

The answer is $2$ . Since $125 = 5^3$ , the equation becomes $5^{2x - 1} = 5^3$ . The base $5$ is positive and not $1$ , so $2x - 1 = 3$ and $x = 2$ . Check that $5^3 = 125$ .

A population is modeled by $P(t) = 2000 \cdot (1.44)^t$ , where $t$ is the number of years. By what percent does the population increase every half year? (A) $12\%$ , (B) $20\%$ , (C) $22\%$ , (D) $44\%$

Show the answer

The correct choice is (B). With $h$ half years, $t = \frac{h}{2}$ , so

\begin{aligned} P &= 2000 \cdot (1.44)^{h/2} \\[0.4em] &= 2000 \cdot (1.2)^h. \end{aligned}

The factor per half year is $\sqrt{1.44} = 1.2$ , a $20\%$ increase. Choice (C) halves the yearly $44\%$ , which ignores compounding, since $1.22^2 = 1.4884$ . Choice (D) is the yearly rate, and (A) misreads $1.2$ as $12\%$ .

Exponential Growth and Decay

Theory, part 69

Practice, part 70

Key terms

Exponential model

Initial value

Growth factor and decay factor

Percent rate

Time interval

Equivalent exponential form

Time-unit conversion

Exponential equations with a common base

Check yourself

How this lesson connects

Builds on

Used later in