Question type

Exponential growth and decay functions on the SAT

An exponential function is $f (t) = a \cdot b^{t}$ : $a$ is the starting amount and $b$ the factor per period. Growth by $r %$ gives $b = 1 + \frac{r}{100}$ ; decay gives $b = 1 - \frac{r}{100}$ . If the change happens every $k$ years, the exponent is $\frac{t}{k}$ : 12% every 5 years is $(0.88)^{t /5}$ .

These questions describe something that grows or shrinks by a percent per period, doubles, halves, or passes through two given values, and ask for the function or a value of it.

It appeared on four of the five official practice tests we analysed, five times in all. The Hard version changes the period, so the exponent becomes a fraction.

On 4 of 5 official practice tests
Desmos helps

Why students miss it

Using the percent as the base. A 10% decrease is a factor of 0.90, not 0.10.

Adding instead of multiplying: $12, 000 - 0.10 n$ is a linear decrease, a different situation.

The period. "Every 5 years" means the factor applies once per 5 years, so the exponent is $\frac{t}{5}$ , not $5 t$ .

How to spot it

"Increases by ...% each year", "decreases by ...% every 5 years", "doubles", "half-life".
Choices of the form $a (b)^{t}$ , some with $t / k$ or $k t$ in the exponent.
Two values of $f$ , such as $f (1) = 12$ and $f (3) = 27$ .

The method

Build $a \cdot b^{t}$

Find the starting amount $a$ , then the factor $b$ for one period.

Grows by $r %$: $b = 1 + r /100$ .
Shrinks by $r %$: $b = 1 - r /100$ .
Once every $k$ units of time: exponent $t / k$ .
Two values given: divide them to find $b$ , then find $a$ .

With Desmos

When two values are given, Desmos finds $a$ and $b$ by regression.

Make a table with the given points, for example (1, 12) and (3, 27).
Type y1~a*b^x1. Desmos returns a and b, and you can evaluate the model at any x.

For a percent-per-period description, writing the function by hand is faster than any graph.

Two real questions

From the 1500prep bank, the way the app shows them. Answer first, then open the explanation.

1Question 1 of 2#381Adv

The value of a machine is $12,000 and decreases by 10% each year. Which function gives the value V of the machine after n years?

Pick an answer, then check it.

Show the answer and explanation

Losing 10% each year leaves 90%, so the yearly multiplier is

0.90

applied to the starting value:

V (n) = 12, 000 (0.90)^{n}

2Question 2 of 2#884Adv

A 640-gram sample of a radioactive material loses 12% of its mass every 5 years. Which function models the mass, in grams, remaining after t years?

Pick an answer, then check it.

Show the answer and explanation

Losing 12% leaves 88%, so the base is

0.88

. That factor applies once per 5-year period, and in

t

years there are

t /5

such periods — so the exponent is

t /5

g (t) = 640 (0.88)^{t /5}

. Check: at

t = 5

you get

640 (0.88) = 563.2

grams.

Questions students ask

How do you write an exponential function from a percent decrease?

Start with the initial amount and multiply by (1 − rate) for each period: a 10% yearly decrease from 12,000 is V(n) = 12,000(0.90)^n.

What does t/5 in the exponent mean?

The change happens once every 5 years. After 5 years the exponent is 1, so the factor has been applied once.

Is exponential growth on the SAT?

Yes, regularly. It appeared on four of the five official digital SAT practice tests we analysed.

Practise this type until it is automatic

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