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Exponential Growth and Decay on the SAT

SAT Math · Advanced Math · Updated October 2026

Before you start

Every exponential model on the SAT has the form $y = a \cdot b^t$. The number $a$ is the starting amount, and $b$ is what the amount is multiplied by each time period.

If $b$ is greater than $1$, the amount grows, and the rate is $b - 1$: a factor of $1.04$ means $4\%$ growth. If $b$ is between $0$ and $1$, it decays, and the rate is $1 - b$: a factor of $0.85$ means $15\%$ decay.

Exponential questions come in a few forms that repeat from test to test. You may need to read the starting value and the rate off a formula, write the formula from a story, or decide whether a situation is linear or exponential. The slip I see most often is mixing up the rate and the factor.

012345 years growth decay

Growth: $y = 10(1.6)^t$, up $60\%$ each year. Decay: $y = 100(0.6)^t$, down $40\%$ each year. Both change by the same factor every year, so neither is a straight line.

Reading the question

You seeDo this
"Increases by $4\%$ each year"Multiply by $1 + 0.04 = 1.04$ each year: $a(1.04)^t$.
"Decreases by $15\%$ each year"Multiply by $1 - 0.15 = 0.85$ each year: $a(0.85)^t$.
"Doubles every $5$ years"$a \cdot 2^{\frac{t}{5}}$. The exponent counts how many $5$-year periods have passed.
"Half-life of $5$ days"$a \cdot \left(\dfrac{1}{2}\right)^{\frac{t}{5}}$. Or count halvings: $15$ days is $3$ halvings.
"Which situation is exponential?"Linear changes by the same amount each step. Exponential changes by the same percent or factor each step.
Two forms of one modelUse exponent rules: $(1.1)^{2t} = \left((1.1)^2\right)^t = (1.21)^t$.

How to work through them

Example 1: writing a decay model

A car is worth \$24,000 and loses $15\%$ of its value each year. Write its value after $t$ years.

  1. Losing $15\%$ leaves $85\%$, so the yearly factor is $0.85$.
  2. The starting value is $24{,}000$.
  3. $V(t) = 24{,}000(0.85)^t$. Using $0.15$ as the factor would mean the car keeps only $15\%$ of its value each year.

Example 2: a growth period longer than one unit

A culture of $50$ bacteria triples every $4$ hours. How many are there after $12$ hours?

  1. The model is $N(h) = 50 \cdot 3^{\frac{h}{4}}$.
  2. After $12$ hours, $\dfrac{12}{4} = 3$ tripling periods have passed.
  3. $N(12) = 50 \cdot 3^3 = 50 \cdot 27 = 1{,}350$.

Example 3: finding $a$ and $b$ from two values

For $f(x) = a \cdot b^x$ with $b > 0$, $f(0) = 5$ and $f(2) = 45$. What is $f(3)$?

  1. $f(0) = a \cdot b^0 = a$, so $a = 5$.
  2. $f(2) = 5b^2 = 45$, so $b^2 = 9$ and $b = 3$.
  3. $f(3) = 5 \cdot 3^3 = 135$.

Example 4: reading a model with a fraction in the exponent

A town's population is modeled by $P(t) = 300(0.9)^{\frac{t}{2}}$, where $t$ is in years. What does the model say?

  1. The factor $0.9$ means a $10\%$ decrease.
  2. The exponent $\dfrac{t}{2}$ goes up by $1$ every $2$ years.
  3. So the population decreases by $10\%$ every $2$ years, starting from $300$.

Common mistakes

Using the rate as the factor. $15\%$ decay is a factor of $0.85$, not $0.15$. $4\%$ growth is $1.04$, not $0.04$ or $4$.

Treating repeated percent changes as simple addition. $6\%$ a year for $10$ years is not $60\%$. It is $(1.06)^{10} \approx 1.79$, about $79\%$.

Ignoring the time period in the exponent. "Every $4$ hours" means the exponent is $\dfrac{h}{4}$, not $h$ or $4h$.

Calling any curve exponential. Look for a constant factor. A table that adds $6$ each time is linear, even if the numbers get big.

Multiplying before applying the exponent. In $3(2)^4$, find $2^4 = 16$ first, then multiply by $3$. It is $48$, not $6^4$.

Ten questions to try

Write down the starting value and the factor before you do anything else.

Easy
The value of an investment, in dollars, after $t$ years is modeled by $V(t) = 500(1.04)^t$. By what percent does the value grow each year?
  • A$0.04\%$
  • B$4\%$
  • C$40\%$
  • D$104\%$
Show solution
Answer: B. The factor is $1.04 = 1 + 0.04$, so the value grows by $0.04 = 4\%$ each year. Choice D is the factor written as a percent: each year's value is $104\%$ of the year before.
EasyStudent-produced response
A population of $200$ insects doubles every year. How many insects are there after $3$ years?
Enter your answer.
Show solution
Answer: $1{,}600$. $200 \cdot 2^3 = 200 \cdot 8 = 1{,}600$. Adding $200$ each year would give only $800$, which is linear growth.
Medium
A car is bought for \$24,000. Its value decreases by $15\%$ each year. Which function gives the value $V$, in dollars, after $t$ years?
  • A$V(t) = 24{,}000(0.15)^t$
  • B$V(t) = 24{,}000 - 0.15t$
  • C$V(t) = 24{,}000(1.15)^t$
  • D$V(t) = 24{,}000(0.85)^t$
Show solution
Answer: D. Each year the car keeps $100\% - 15\% = 85\%$ of its value, so the factor is $0.85$. Choice A keeps only $15\%$ each year, and choice C is a $15\%$ increase.
Medium
A culture starts with $50$ bacteria, and the number triples every $4$ hours. Which function gives the number of bacteria $h$ hours after the start?
  • A$N(h) = 50 \cdot 3^{4h}$
  • B$N(h) = 50 \cdot 3^{\frac{h}{4}}$
  • C$N(h) = 50 \cdot 4^{\frac{h}{3}}$
  • D$N(h) = 150^{\frac{h}{4}}$
Show solution
Answer: B. The amount is multiplied by $3$ once every $4$ hours, so the number of triplings after $h$ hours is $\dfrac{h}{4}$. Check: after $4$ hours, $50 \cdot 3^1 = 150$. Choice A would triple the culture $4$ times every hour.
MediumStudent-produced response
For the function $f(x) = 3(2)^x$, what is the value of $f(4) - f(3)$?
Enter your answer.
Show solution
Answer: $24$. $f(4) = 3 \cdot 16 = 48$ and $f(3) = 3 \cdot 8 = 24$. The difference is $48 - 24 = 24$.
Medium
A sample of a radioactive substance has a mass of $80$ grams. Its half-life is $5$ days. What is the mass of the sample after $15$ days?
  • A$40$ grams
  • B$20$ grams
  • C$10$ grams
  • D$5$ grams
Show solution
Answer: C. $15$ days is $\dfrac{15}{5} = 3$ half-lives. $80 \to 40 \to 20 \to 10$. So $10$ grams remain. Choice A stops after one half-life.
Hard
Which expression is equivalent to $1{,}000(1.21)^t$?
  • A$1{,}000(1.1)^{2t}$
  • B$1{,}000(1.105)^{2t}$
  • C$1{,}000(1.21)^{2t}$
  • D$1{,}000(1.1)^{\frac{t}{2}}$
Show solution
Answer: A. $(1.1)^2 = 1.21$, so $(1.1)^{2t} = \left((1.1)^2\right)^t = (1.21)^t$. Choice B splits the $21\%$ in half, but compounding does not work that way: $(1.105)^2 \approx 1.221$.
Hard
An investment of \$2,500 grows by $6\%$ each year. To the nearest dollar, what is its value after $10$ years?
  • A\$4,000
  • B\$4,477
  • C\$2,650
  • D\$26,500
Show solution
Answer: B. $2{,}500(1.06)^{10} \approx 2{,}500 \cdot 1.7908 \approx 4{,}477$. Choice A adds $6\%$ of the original amount every year, which is simple interest. Choice C is the value after one year.
HardStudent-produced response
The function $f$ is defined by $f(x) = a \cdot b^x$, where $a$ and $b$ are positive constants. If $f(0) = 5$ and $f(2) = 45$, what is the value of $f(3)$?
Enter your answer.
Show solution
Answer: $135$. $f(0) = a = 5$. Then $5b^2 = 45$, so $b^2 = 9$ and $b = 3$. $f(3) = 5 \cdot 27 = 135$.
Hard
The population of a town is modeled by $P(t) = 300(0.9)^{\frac{t}{2}}$, where $t$ is the number of years since 2020. Which statement is true according to the model?
  • AThe population decreases by $10\%$ every year.
  • BThe population decreases by $5\%$ every year.
  • CThe population decreases by $10\%$ every $2$ years.
  • DThe population decreases by $90\%$ every $2$ years.
Show solution
Answer: C. The factor $0.9$ is a $10\%$ decrease, and the exponent $\dfrac{t}{2}$ increases by $1$ every $2$ years. So the population drops by $10\%$ every $2$ years. Choice B halves the rate, but two drops of $5\%$ leave $0.95^2 = 0.9025$, not $0.9$.

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