SAT exponential growth and decay, step by step

The key idea

f(t) = a(1 + r)^t \qquad f(t) = a(1 - r)^t \qquad f(t) = a \cdot 2^{\frac{t}{d}}

The first is growth by rate r per step, the second is decay by rate r per step (r as a decimal), and the third doubles every d units of time. Dividing t by d counts how many doubling periods have passed.

Linear or exponential?
	Linear	Exponential
Each step	Adds the same amount	Multiplies by the same factor
Table test	Equal differences	Equal ratios
Example	$5, 8, 11, 14$ (add 3)	$5, 10, 20, 40$ (times 2)

Worked examples

Example 1: growth by a percent

Problem A savings account starts with $2,000 and grows 5% per year. How much is in the account after 3 years?

Growth of 5% means a factor of 1.05 each year.
$A = 2000(1.05)^3$
$1.05^3 = 1.157625$ .
$2000 \times 1.157625 = 2315.25$

Answer $2,315.25

Example 2: build the model from a table

Problem A table shows $f(0) = 400$ , $f(1) = 300$ , $f(2) = 225$ , $f(3) = 168.75$ . Write f and describe the change.

Check ratios, not differences. Each output divided by the one before:
$\frac{300}{400} = \frac{225}{300} = 0.75$
The ratio is constant, so f is exponential with factor 0.75 and start 400.
$f(x) = 400(0.75)^x$
A factor of 0.75 keeps 75% each step, which is a 25% decrease.

Answer $f(x) = 400(0.75)^x$ : it decreases by 25% each step.

Example 3: doubling time

Problem A bacteria population starts at 500 and doubles every 4 hours. What is the population after 1 day?

One day is 24 hours, which is 6 doubling periods.
$\frac{24}{4} = 6$
Multiply by 2 six times.
$500 \cdot 2^6 = 32000$

Answer 32,000

Example 4 (SAT-hard): change the time unit

Problem A population is modeled by $P(t) = 1200(1.21)^{\frac{t}{2}}$ , where t is in years. By what percent does the population grow each year?

Rewrite the power: $(1.21)^{\frac{t}{2}} = \left((1.21)^{\frac{1}{2}}\right)^t$ .
$(1.21)^{\frac{1}{2}} = \sqrt{1.21} = 1.1$
So $P(t) = 1200(1.1)^t$ . A yearly factor of 1.1 means 10% growth per year.
The trap answer is 21%, which is the growth every 2 years. Over 2 years: $1.1 \times 1.1 = 1.21$ , so it matches.

Answer 10% per year

Common mistakes

Using the rate as the factor. A 15% decay is a factor of 0.85, not 0.15. Fix: factor equals 1 plus or minus the rate.
Calling a constant difference exponential. If a table adds the same number each step, it is linear. Fix: check ratios for exponential, differences for linear.
Forgetting to divide the time. "Doubles every 4 hours" uses $2^{\frac{t}{4}}$ , not $2^{4t}$ . Fix: ask how many periods fit in the time given.
Multiplying the percent by the number of years. 5% a year for 3 years is not 15%: $1.05^3 \approx 1.158$ , so it is about 15.8%. Fix: use the exponent.
Reading 21% as the yearly rate in a model with $\frac{t}{2}$ . The factor 1.21 applies every 2 years. Fix: rewrite the base so the exponent is t.

Quick method

Practice

5 SAT-style questions

Which statement describes $f(x) = 80(0.92)^x$ ?
1. It increases by 92% each time x increases by 1.
2. It decreases by 8% each time x increases by 1.
3. It decreases by 92% each time x increases by 1.
4. It decreases by 0.08 each time x increases by 1.
Show answer
Answer: It decreases by 8% each time x increases by 1.
The factor 0.92 keeps 92% of the value each step, so 8% is lost. 0.08 is a percent of the value, not a fixed amount, so the last choice describes a linear function.
A car worth $24,000 loses 15% of its value each year. Which expression gives its value after t years?
1. $24000(0.15)^t$
2. $24000(0.85)^t$
3. $24000(1.15)^t$
4. $24000 - 0.15t$
Show answer
Answer: $24000(0.85)^t$
Losing 15% keeps 85%, so the factor is 0.85. Using 0.15 would keep only 15% each year. 1.15 is growth. The last choice subtracts 15 cents a year.
A sample of 3,000 cells triples every 5 days. How many cells are there after 15 days?
1. 9,000
2. 27,000
3. 45,000
4. 81,000
Show answer
Answer: 81,000
15 days is 3 tripling periods: $3000 \cdot 3^3 = 81{,}000$ . 9,000 triples once, and 45,000 multiplies 3,000 by 15.
The value of an account t years from now is $V(t) = 5000(1.04)^t$ . Which expression gives the value m months from now?
1. $5000(1.04)^{12m}$
2. $5000(1.04)^{\frac{m}{12}}$
3. $5000\left(\frac{1.04}{12}\right)^m$
4. $5000(1.0033)^m$
Show answer
Answer: $5000(1.04)^{\frac{m}{12}}$
m months is $\frac{m}{12}$ years, so replace t with $\frac{m}{12}$ . 1.0033 is a rounded guess, and $1.04^{\frac{1}{12}}$ is about 1.00327, so it is not equal.
Student-produced response: $f(x) = a \cdot b^x$ , where a and b are positive constants. If $f(0) = 6$ and $f(2) = 54$ , what is $f(3)$ ?
Show answer
Answer: 162
$f(0) = a = 6$ . Then $6b^2 = 54$ , so $b^2 = 9$ and $b = 3$ (b is positive). $f(3) = 6 \cdot 27 = 162$ .

Frequently asked questions

How do I tell exponential from linear in a table?

Look at how the outputs change when x goes up by 1. Equal differences (add 3 each time) mean linear. Equal ratios (multiply by 2 each time) mean exponential. Make sure the x values are evenly spaced before you compare.

What is the difference between growth rate and growth factor?

The rate is the percent change, like 5%. The factor is what you multiply by each step, like 1.05. For decay, a rate of 15% gives a factor of 0.85. The factor goes in the base of the exponential.

Are exponential questions hard on the SAT?

Basic ones are quick. The harder ones change the time unit, like a model in years asked about months, or hide the rate inside an exponent like $\frac{t}{2}$ . Rewriting the base so the exponent is plain t solves most of them.

Sources

College Board: SAT Math, Advanced Math skills (nonlinear functions), accessed October 1, 2026

Exponential growth and decay on the SAT

The key idea

Worked examples

Common mistakes

Quick method

Practice

Frequently asked questions

Sources

Try asking Ducky

The key idea

Worked examples

Common mistakes

Quick method

Practice

Frequently asked questions

Sources

Related

Try asking Ducky