Exponential growth and decay questions model anything that changes by a percentage rather than a fixed amount — population, investments, radioactive decay. This guide covers the formula the SAT expects you to know, shows the difference between growth and decay, and walks through two full worked examples — one growth, one decay.
The Exponential Model the SAT Uses
Every exponential function on the SAT follows the same basic shape:
\(f(x) = a \cdot b^x\)
Here \(a\) is the starting value (what you have at \(x=0\)), and \(b\) is the growth or decay factor. When \(b > 1\), the quantity grows; when \(0 < b < 1\), it decays. This is one of the function families covered under nonlinear functions, and it's worth memorizing since the reference sheet gives you no help with exponential identities.
Growth vs. Decay: Reading the Factor
A percentage change translates into the growth factor \(b\) like this:
- A quantity that grows by \(r\%\) each period has \(b = 1 + \frac{r}{100}\).
- A quantity that decays (shrinks) by \(r\%\) each period has \(b = 1 - \frac{r}{100}\).
- The larger \(b\) is above 1, the faster the growth; the closer \(b\) is to 0, the faster the decay.
| Scenario | Growth factor \(b\) | Type |
|---|---|---|
| Increases 8% per year | \(1.08\) | Growth |
| Decreases 8% per year | \(0.92\) | Decay |
| Doubles every period | \(2\) | Growth |
| Halves every period | \(0.5\) | Decay |
| Decreases 2.5% per month | \(0.975\) | Decay |
Worked Example: Compound Growth
An investment starts at $2,000 and grows 5% each year. Write the function and find its value after 6 years.
- Identify the starting value: \(a = 2000\).
- Convert the percentage to a growth factor: \(b = 1 + \frac{5}{100} = 1.05\).
- Write the function: \(f(t) = 2000(1.05)^t\), where \(t\) is years.
- Substitute \(t = 6\): \(f(6) = 2000(1.05)^6\).
- Compute \((1.05)^6 \approx 1.3401\).
- Multiply: \(f(6) = 2000 \times 1.3401 \approx 2680.19\).
- After 6 years, the investment is worth approximately \(\$2,680\).
The embedded Desmos calculator can compute a power like \((1.05)^6\) instantly, so don't waste time doing it by hand — type the whole expression in and read off the result.
Worked Example: Half-Life Decay
A radioactive sample weighs 80 grams and has a half-life of 3 hours, meaning its mass is cut in half every 3 hours. Write a function for its mass and find how much remains after 12 hours.
- Identify the starting value: \(a = 80\).
- Since the mass halves every 3 hours, the decay factor per 3-hour period is \(b=0.5\). Write the exponent in terms of the number of 3-hour periods that have passed: \(f(t) = 80(0.5)^{t/3}\), where \(t\) is time in hours.
- Substitute \(t=12\): \(f(12) = 80(0.5)^{12/3} = 80(0.5)^4\).
- Compute \((0.5)^4 = 0.0625\).
- Multiply: \(f(12) = 80 \times 0.0625 = 5\).
- After 12 hours — exactly four half-lives — 5 grams remain.
Notice the exponent is written as \(t/3\), not just \(t\), because the "5% each year" style problem changes every single unit of \(t\), while a half-life problem changes only every fixed period. Matching the exponent's units to the period described is the step students most often get wrong on decay problems.
Spotting Exponential Situations in Word Problems
The SAT rarely says "this is exponential" directly. Instead, look for these signals:
- Phrases like "increases by a factor of," "doubles every," "decreases by X% each," or "half-life."
- A table where each output is a constant multiple of the previous one, not a constant difference.
- Any context involving compounding — interest, population, bacteria cultures, radioactive material.
If the change described is a fixed percentage rather than a fixed amount, it's exponential, not linear — this distinction alone resolves a large share of "which equation models this" questions.
Comparing Exponential and Quadratic Growth
Both curve, but they behave very differently. A quadratic function's rate of change grows linearly, while an exponential function's rate of change grows exponentially — over a long enough interval, an exponential function will always eventually overtake a quadratic one, no matter how large the quadratic's coefficients are. This distinction occasionally shows up in comparison questions asking which model better fits a data set over many periods.
Grid-In Questions and Exponential Functions
When an exponential problem appears as a grid-in rather than multiple choice, the SAT typically asks for a specific numeric value at a given time, or for the value of \(a\) or \(b\) itself given two data points. In that second case, divide one output by another to isolate \(b\): if \(f(1)=210\) and \(f(0)=200\), then \(b = \frac{210}{200}=1.05\). Keep in mind that a grid-in answer built from a decayed or grown value is often not a whole number, so round only if the question tells you to, and enter as much precision as the grid allows otherwise.
Building This Skill With Practice
Exponential growth and decay problems reward pattern recognition more than raw computation, which makes timed repetition especially effective. Our instant-scored SAT Math sets include Nonlinear Functions practice at Easy, Medium, and Hard levels, each around 22 questions, so you can build speed recognizing growth-factor language before you see it on the real test. If function notation itself — reading \(f(x)\), transformations, and composite functions — feels shaky, our function notation and transformations guide covers that foundation directly.
Frequently asked questions
Is exponential decay ever written with a negative exponent?
Sometimes, though it's mathematically equivalent to using a growth factor between 0 and 1. Either way, remember \(b^{-1} = \frac{1}{b}\) since exponent rules aren't on the reference sheet.
How do I know if a word problem is exponential rather than quadratic?
Look for percentage-based or "doubling/halving" language. Quadratic contexts usually describe area, projectile motion, or a rate that changes by a constant amount each step, not a constant ratio.
Do I need to know the natural exponential base \(e\)?
It's uncommon on the digital SAT compared to the \(a \cdot b^x\) form. Focus your memorization on percentage-to-growth-factor conversion first.
Why does the exponent in a half-life function use \(t/3\) instead of just \(t\)?
Because the decay factor of \(0.5\) applies once per full half-life period, not once per single unit of time. Dividing \(t\) by the length of that period converts elapsed time into "number of half-lives passed," which is what the exponent actually needs to represent.