SAT Math skill page

SAT Exponential Functions Practice

Learn how exponential growth and decay show up in SAT tables, equations, and word problems.

Practice exponential functions in Dolphin SAT Practice questions

12-18 min practice time3 questions on pageAdvanced Math

Practice time12-18 min

Question bank3 questions

Best forAdvanced Math

What this tests

What to know for this SAT skill

Recognize exponential growth and decay from equations
Interpret the starting value and growth factor
Compare linear and exponential patterns
Use tables to find a constant multiplier

Dolphin question bank

Practice your skill

Question 1 of 3Hard

P(t) = 260(

1.04)^{\frac{t}{4}}

The function P models the population, in thousands, of a certain city t years after 2003. According to the model, the population is predicted to increase by 4% every n months. What is the value of n?

Show answer and solution

Correct answerA. 48

Solution walkthrough

In the model $P (t) = 260 (1.04)^{t /4}$ , the factor $1.04$ means the population increases by 4% whenever the exponent increases by 1.
Since the exponent is $t /4$ , it increases by 1 when $t$ increases by 4 years.
Four years is 4 times 12 = 48 months, so $n = 48$ .

Why the other choices miss

Choice B is incorrect because 12 months is 1 year, not 4 years. Choice C is incorrect because 18 months does not make the exponent increase by 1. Choice D is incorrect because 72 months is 6 years, not 4 years.

Question 2 of 3Hard

The population of a town is currently 50,000, and the population is estimated to increase each year by 3% from the previous year. Which of the following equations can be used to estimate the number of years, t, it will take for the population of the town to reach 60,000?

$50, 000 = 60, 000 (0.03)^{t}$
$50, 000 = 60, 000 (3)^{t}$
$60, 000 = 50, 000 (0.03)^{t}$
$60, 000 = 50, 000 (1.03)^{t}$

Show answer and solution

Correct answer $D .60, 000 = 50, 000 (1.03)^{t}$

Solution walkthrough

Stating that the population will increase each year by 3% from the previous year is equivalent to saying that the population each year will be 103% of the population the year before.
Since the initial population is 50,000, the population after $t$ years is given by $50, 000 (1.03)^{t}$ .
It follows that the equation $60, 000 = 50, 000 (1.03)^{t}$ can be used to estimate the number of years it will take for the population to reach 60,000.

Why the other choices miss

Choice A is incorrect. This equation models how long it will take the population to decrease from 60,000 to 50,000, which is impossible given the growth factor. Choice B is incorrect and may result from misinterpreting a 3% growth as growth by a factor of 3. Additionally, this equation attempts to model how long it will take the population to decrease from 60,000 to 50,000. Choice C is incorrect and may result from misunderstanding how to model percent growth by multiplying the initial amount by a factor greater than 1.

Question 3 of 3Hard

For the function f, f(0) = 86, and for each increase in x by 1, the value of f(x) decreases by 80%. What is the value of f(2)?

3.44
17.2
68.8
0.344

Show answer and solution

Correct answerA. 3.44

Solution walkthrough

The correct answer is 3.44. It's given that $f (0) = 86$ and that for each increase in $x$ by 1, the value of $f (x)$ decreases by 80%.
Because the output of the function decreases by a constant percentage for each 1-unit increase in the value of $x$ , this relationship can be represented by an exponential function of the form $f (x) = a (b)^{x}$ , where $a$ represents the initial value of the function and $b$ represents the rate of decay, expressed as a decimal. Because $f (0) = 86$ , the value of $a$ must be 86.
Because the value of $f (x)$ decreases by 80% for each 1-unit increase in $x$ , the value of $b$ must be $(1 - 0.80)$ , or 0.2. Therefore, the function $f$ can be defined by $f (x) = 86 (0.2)^{x}$ .
Substituting 2 for $x$ in this function yields $f (2) = 86 (0.2)^{2}$ , which is equivalent to $f (2) = 86 (0.04)$ , or $f (2) = 3.44$ . Either 3.44 or 86/25 may be entered as the correct answer.

Avoid these traps

Common mistakes on this skill

Treating growth as repeated addition

Exponential change uses a constant multiplier over equal intervals, not a constant amount added each time.

Confusing the initial value and growth factor

In a model such as a(b)^x, a is the value at x = 0 and b controls growth or decay.

Study plan

How to practice this skill in Dolphin

Identify the initial value and the equal time interval in the problem.
Convert the percent change into a multiplier such as 1.08 or 0.92.
Check the model at x = 0 and after one interval before interpreting the requested value.

Practice exponential functions in Dolphin SAT

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FAQ

Questions about SAT Exponential Functions Practice

How can I recognize an exponential relationship?

Look for equal inputs producing equal ratios between outputs. A linear relationship instead produces equal differences.

How is exponential decay represented?

The multiplier is between 0 and 1. A 12 percent decrease, for example, uses a multiplier of 0.88.