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Exponential Growth and Decay on the SAT

The reader has run into an SAT math question about exponential growth or decay and wants a reliable way to turn the word problem into an equation and solve it.

4 min readReviewed October 21, 2026StudyPebble
Next SAT November 7, 202617 days awayregister by October 23, 2026

How to tell it's an exponential question

Linear change adds the same amount every time. Exponential change multiplies by the same factor every time. That distinction is the whole question, and the SAT usually tells you which one you're in with its wording. "Increases by 40 each month" is linear. "Increases by 8% each month" is exponential. Spot the percent, and you've spotted the pattern.

Watch for phrases like "each year," "every hour," "doubles every," "decreases by," or "half-life." A one-time change (a coupon that knocks 20% off a single price) isn't this kind of question at all. Repeated change over equal time steps is the signal you want.

Try one
Nonlinear functions
Real question
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The graph of the quadratic function f is shown in the xy-plane. What is the y-coordinate of the vertex of the graph?

Building the equation: y = a(b)^x

Start by pulling out two numbers: a, the value at the very start, and the percent change per period. Everything else is plugging in. If a bacteria culture starts at 800 cells and grows 12% every hour, a is 800 and the growth factor is 1 + 0.12, or 1.12. The equation is y = 800(1.12)^x, with x in hours.

Want the population after 5 hours? Plug in 5: 800(1.12)^5 works out to about 1,410 cells. Decay works the same way but subtracts. A car bought for $24,000 that loses 15% of its value each year gives a factor of 1 minus 0.15, or 0.85, so the equation is y = 24000(0.85)^x. After 4 years, that's 24000(0.85)^4, around $12,528.

The trap here is small but costs real points: writing the decay factor as 0.15 instead of 0.85. A base of 0.15 says the car loses 85% of its value a year, which isn't the question being asked. Always convert the percent that's left over, not the percent that's gone.

Half-life questions describe a quantity that cuts in half every fixed stretch of time: a drug leaving the bloodstream, a radioactive sample decaying, a signal fading.

Half-life and repeated-halving problems

Half-life questions describe a quantity that cuts in half every fixed stretch of time: a drug leaving the bloodstream, a radioactive sample decaying, a signal fading. The form is still y = a(0.5)^(t/h), where h is the half-life and t is however much time has passed. The exponent isn't just t. It's t divided by h, because that ratio tells you how many halvings actually happened.

Say a dose of 240 mg drops by half every 6 hours. After 18 hours, how much is left? Eighteen divided by six is three, so three halvings have happened: 240(0.5)^3. That's 240 times one eighth, or 30 mg. Notice you never needed a calculator for the exponent itself, since counting half-lives is just division.

The same structure covers doubling, too. A population that doubles every 3 days is y = a(2)^(t/3). Same logic, opposite direction: the base flips from 0.5 to 2, nothing else changes.

When the SAT hands you the equation instead

A lot of these questions skip the word problem entirely. They just hand you an equation and ask what a number inside it means in context. Given y = 1500(0.94)^x for the value of equipment after x years, a question might ask what 1500 represents, or what 0.94 tells you about the rate of decline.

Read the base as a percent change, not as a raw decimal. A factor of 0.94 means 94% of the value carries over each year, which means the equipment loses 6% a year, not 94%. That one-sentence habit of translating b into "loses (1 minus b) percent" or "gains (b minus 1) percent" clears up most of these interpretation questions in a few seconds.

Multiple choice versions often swap in a close but wrong equation, like using 1.06 instead of 0.94, or attaching the percent to the wrong variable. Plugging in x = 0 is a fast check: it should return the starting value exactly, with nothing added or multiplied beyond a.

Common traps worth checking twice

Mixing up which direction the base points causes most of the wrong answers here. Before you commit to a base, ask in plain words: is this thing growing or shrinking? Then check that your base is above or below 1 to match. A base over 1 for something that's supposed to be decaying means you added the rate instead of subtracting it.

Another common slip is treating a half-life exponent as plain elapsed time instead of elapsed time over the half-life. If the problem gives you 18 hours and a 6-hour half-life, the exponent is 3, not 18. Write out the division before you touch the exponent and this one mostly takes care of itself.

For a quick gut check on multiple choice, StudyPebble's practice by skill groups these under nonlinear functions with a worked explanation on every question, which helps if a particular setup, like percent decrease versus percent remaining, keeps tripping you up.

Questions students ask

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What's the difference between linear and exponential growth on the SAT?

Linear growth adds a fixed amount each period, like $50 a month. Exponential growth multiplies by a fixed factor each period, like 5% a month. If the problem names a percent rate repeated over time, it's exponential, not linear.

How do you turn a percent decrease into the base of an exponential equation?

Subtract the percent, written as a decimal, from 1. A 20% decrease gives a base of 0.8, since 80% of the value remains after each period. Using 0.2 as the base is the single most common mistake on these questions.

Does the SAT ever test half-life directly?

Yes, usually dressed up as a medication, a radioactive sample, or a quantity that cuts in half on a fixed schedule. The setup is y = a(0.5)^(t divided by the half-life), where dividing elapsed time by the half-life gives the number of halvings.

Can Desmos handle exponential growth and decay questions?

Yes. Desmos is available for every question in both Math modules, so you can type the equation in directly and graph it, trace a specific x-value, or just use it to evaluate a power instead of computing by hand.

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