Interpreting Exponential Growth and Decay
Exponential models look like $y = a \cdot b^x$. The base $b$ is the growth factor; $a$ is the starting amount.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
A growth factor of $1.05$ means a $5\%$ increase per step ($b = 1 + r$). The SAT tests whether you can separate the factor from the rate, and the rate from the starting value $a$.
How to solve one, step by step
Example: $y = 200(1.05)^t$.
- Starting amount $a = 200$.
- Growth factor $1.05$ means $5\%$ growth per year.
The mistakes that cost points
- Reading the factor as the rate. $1.05$ is a $5\%$ rate, not $105\%$ growth.
- Confusing the growth rate with the starting value. $a$ is the initial amount; $b$ controls the change.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
In the equation $A = 25{,}000(1.60)^t$, $t$ represents the number of time periods. Which of the following correctly interprets the equation?
- AAn initial amount of 25,000 grows at a rate of 160% per time period.
- BAn initial amount of 25,000 grows at a rate of 60% per time period.
- CAn initial amount of 1.60 grows by a factor of 25,000 per time period.
- DAn initial amount of 25,000 grows at a rate of 1.60 per time period.
Show solution
Answer: B, An initial amount of 25,000 grows at a rate of 60% per time period.. For an exponential model in the form $A = P(1 \pm r)^t$, the number outside the parentheses ($P$) is the initial amount, and the value of $r$ in the parentheses tells the growth rate. Convert the factor to a percent: 1.60 means the quantity grows by 60% per period.
Medium
A radioactive substance decays according to $A(t) = 200 \cdot (k)^t$, where $A$ is the amount remaining and $t$ is time in years. After 1 year, $180$ grams remain. What is the value of $k$?
- A$k = 0.9$
- B$k = 1.9$
- C$k = 90$
- D$k = 200$
Show solution
Answer: A, $k = 0.9$. $A(1) = 200 \cdot k^1 = 200k = 180 \Rightarrow k = 0.9$. Since $k < 1$, this confirms decay. The decay rate is $1 - 0.9 = 10\%$ per year.
Hard
A colony of bacteria doubles every hour according to $N(t) = N_0 \cdot 2^t$. A different model is written as $N(t) = 50 \cdot (1 + k)^t$, where $k$ is the hourly growth rate as a decimal. If the two models are equivalent and $N_0 = 50$, what is the value of $k$?
- A$k = 50$
- B$k = 1$
- C$k = 2$
- D$k = -1$
Show solution
Answer: B, $k = 1$. Since doubling means $b = 2$, and $1 + k = 2$, we have $k = 1$. The growth rate is $100\%$ per hour (doubling = 100% increase). A common error is saying $k = 2$ (using the factor instead of the rate).
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