Build, identify, and interpret exponential models

Your feedback matters:

Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 30 minutes
Domains: Advanced Math
Techniques: Exponential-modelsGrowth-decayConstant-ratioInterval-factorMethod-choice

What you’ll learn

Tell a pattern that adds the same amount each step from one that multiplies by the same factor.
Write an exponential model from its starting value, its factor and its interval, and say what each part means.
Turn a percent increase or decrease into a factor.
Tell the part that remains apart from the percent that’s lost.
Change a factor to a different interval.

Prerequisites

Why this matters on the SAT

Recognize growth, then build the right model

Some quantities grow or shrink by the same factor over and over, once every equal interval. That’s exponential change. An SAT question might tell it as a story, show it in a table or hand you the equation, but your job is the same each time. Find three things: where it starts, what it’s multiplied by, and how often that happens. Start, factor, interval.

Once you have those, you can build the model, say what its factor means and change the factor to a new interval. Don’t treat the change as linear, though. A line adds the same amount each time, and this multiplies.

Solution to the example

Two things decide this one: the full growth factor, and an exponent that counts four-day stretches.

A $15\%$ increase keeps the whole population, $100\%$ , and adds $15\%$ more. So each new population is $115\%$ of the one before, which is $1.15$ times as big.

Now the exponent. The colony doesn’t grow by $15\%$ every day. It grows by $15\%$ every $4$ days, so the exponent has to count four-day intervals, and $\frac{t}{4}$ does exactly that. At $t=8$ , for example, $\frac{8}{4}=2$ : two intervals, so two rounds of growth. The answer is B.

Each wrong choice makes a classic slip. A grows the colony every day instead of every $4$ days. C keeps only $15\%$ of the population each interval. D adds the same amount each time instead of multiplying the new, bigger population.

SAT example

A colony starts with $320$ insects and increases by $15\%$ every $4$ days. Which function models the population $P(t)$ after $t$ days?

A
$P(t)=320(1.15)^t$
B
$P(t)=320(1.15)^{t/4}$
C
$P(t)=320(0.15)^{t/4}$
D
$P(t)=320\left(1+0.15\frac{t}{4}\right)$

Recognize the ratio

Picture a table where the input goes up in equal steps. Here, each step is $2$ hours. A linear pattern adds the same amount every step. An exponential pattern multiplies by the same factor every step. Which one is this?

Hours	$0$	$2$	$4$	$6$
Culture size	$90$	$135$	$202.5$	$303.75$

Start by subtracting each output from the next one. These differences aren’t the same:

45,\quad 67.5,\quad 101.25.

Now divide each output by the one before it. These ratios are all the same:

\frac{135}{90} = \frac{202.5}{135} = \frac{303.75}{202.5} =1.5.

A ratio of $1.5$ means each culture size is $1.5$ times the one before. That same multiplication repeats every $2$ hours, so the relationship is exponential. In short: same difference, linear; same ratio, exponential.

Check exact ratios first whenever a table follows a clean pattern. This one does, so it doesn’t need regression. Real measurements can be messier. If they vary and the ratios are close but not exactly equal, don’t pick one pair and treat its ratio as exact. A noisy table like that calls for exponential regression, covered in the lines of best fit lesson.

Check your understanding:

A quantity goes from $40$ to $52$ to $64$ to $76$ over equal intervals. Is the pattern linear or exponential, and how can you tell?

Common mistake:

Calling every growing pattern exponential. Going up doesn’t make a pattern exponential; how it goes up does. Compare each output with the next one over equal input steps. The same difference means linear, and the same ratio means exponential. Before you decide, check every consecutive pair, not only the first.

Write the model

Say a population starts at $750$ and doubles every $6$ hours. It has doubled once by hour $6$ , twice by hour $12$ , and so on, so the exponent has to count six-hour stretches. That’s $\frac{t}{6}$ :

P(t)=750(2)^{t/6},

where $t$ is time in hours and $P(t)$ is the population after $t$ hours. At $t=0$ , the exponent is $0$ , so $P(0)=750$ , the starting value. At $t=6$ , one full interval has passed, the exponent is $1$ , and $P(6)=1500$ .

In general, once you see repeated multiplication, you can write it as

y=a\,b^{t/k},

where:

$a$ is the starting value, at $t=0$ (the $750$ above)
$b$ is the factor, the number you multiply by, for one full interval (the $2$ )
$k$ is how long one interval is (the $6$ hours)
$t$ is how much of the input has passed, such as the time so far
$\frac{t}{k}$ is how many intervals have passed, whole or partial

Dividing $t$ by $k$ tells you how many intervals fit into it. When $t=k$ , the exponent is $1$ . When $t=2k$ , it’s $2$ .

So a sentence in words turns straight into the model:

\text{starts at }a\text{ and multiplies by }b\text{ every }k\text{ units} \quad\longrightarrow\quad a\,b^{t/k}.

If a question gives you the model and asks which input gives a certain output, you’re solving an equation instead, as in Solve exponential equations.

Turn a percent into a factor

Here, $p$ is the percent number, so $p\%=\frac{p}{100}$ . Think of the whole starting amount as $1$ . An increase keeps all of it and adds $\frac{p}{100}$ more, so

b=1+\frac{p}{100}.

A decrease takes away $\frac{p}{100}$ and keeps the rest, so

b=1-\frac{p}{100}.

For example:

\begin{aligned} 18\%\text{ increase}&\longrightarrow 1+0.18=1.18,\\[1.4em] 18\%\text{ decrease}&\longrightarrow 1-0.18=0.82. \end{aligned}

Don’t drop the $1$ . It’s everything you already had, and forgetting it is exactly how choice C in the SAT example went wrong.

Check your understanding:

A machine is worth $24{,}000$ dollars and loses $7\%$ of its value every year. Write a model for its value $V(t)$ after $t$ years.

Read every number

Take the model

m(t)=80(0.72)^{t/6},

where $t$ is time and $m(t)$ is the amount at that time. Try putting each part into words:

It starts at $80$ .
The $\frac{t}{6}$ counts six-unit intervals, so the exponent goes up by $1$ each time $t$ goes up by $6$ .
Every $6$ time units, the amount is multiplied by $0.72$ .
So every $6$ time units, $72\%$ of the previous amount remains and $28\%$ of it is lost.

Here’s the trap: $0.72$ doesn’t mean a $72\%$ decrease. The factor is what’s left. To find what’s lost, ask how far the factor is below $1$ :

1-0.72=0.28.

Check your understanding:

The function $q(t)=150(0.35)^{t/4}$ models a quantity after $t$ days. During each four-day interval, what percent of the previous amount remains, and what percent is lost?

Common mistake:

Treating $b$ as the factor for every single unit. The $\frac{t}{k}$ in the exponent sets the interval: $b$ is applied once every $k$ units, each time the exponent goes up by $1$ . To check your reading, plug in $t=k$ . The model becomes $a\,b$ , so the factor has been used exactly once.

Example: Change the interval with Desmos

Worked example

The mass $M(t)$ , in milligrams, of a medication in the bloodstream $t$ hours after a dose is modeled by

M(t)=500(0.4096)^{t/20}.

The mass decreases by $p\%$ of its previous value every $5$ hours. What is the value of $p$ ?

Step 1

Find the factor and its interval

The factor $0.4096$ applies every $20$ hours, because the exponent $\frac{t}{20}$ goes up by $1$ each time $t$ goes up by $20$ .

Step 2

Build the five-hour factor

Five hours is $\frac{5}{20}=\frac14$ of the $20$ -hour interval. Call the five-hour factor $c$ . Four five-hour stretches make up one $20$ -hour interval, so applying $c$ four times has to give the $20$ -hour factor:

c^4=0.4096.

Take the positive fourth root, since a factor for a positive amount can’t be negative:

c =0.4096^{1/4} =0.4096^{5/20}.

Look at that last exponent: it’s the new interval over the old one, $\frac{5}{20}$ .

But the question asks for the percent lost, not the five-hour factor. So let Desmos do the whole job in one expression:

(1-0.4096^(5/20))*100

Desmos shows $20$ . That one expression changes the interval, finds the part that’s lost and turns it into a percent, and you never round anything along the way.

Calculator loads as you approach

The percent lost every five hours, in one expression.

Step 3

Read the percent lost

A result of $20$ means the mass drops by $20\%$ every five hours. Put another way, the five-hour factor is $0.8$ , so $80\%$ remains. So

\boxed{p=20}.

Try it yourself:

Predict first: will the $10$ -hour factor be greater or less than $0.8$ ? Think about why, knowing the medication keeps decreasing. Now change 5/20 to 10/20 in the calculator. The result is the percent lost, so $100$ minus it is the percent that remains. Was your prediction right? Reset the calculator when you’re done.

This is the tricky part of interval questions. Each percent loss comes out of the updated amount, which keeps shrinking, so you can’t multiply the percent by the number of intervals. Four $20\%$ losses aren’t an $80\%$ loss: they leave $0.8^4=0.4096$ of the dose, so about $59\%$ is lost over $20$ hours. Change the factor with a power or a root instead.

Related: For optional, more advanced calculator work with exponential forms, targets and Log Mode, see Exponential models, transformations, and Log Mode.

Practice problems

For each one, find the start, the factor and the interval before you calculate.

Build a model from a table

Practice problem

The table shows values of an exponential function $F$ .

$t$	$0$	$4$	$8$
$F(t)$	$180$	$270$	$405$

Which function represents $F(t)$ ?

Calculator loads as you approach

Use Desmos if it helps you solve or check this problem.

Read the interval factor

Practice problem

The mass $M(t)$ , in grams, of a sample $t$ days after an experiment begins is modeled by

M(t)=40(0.65)^{t/8}.

Which statement best explains the factor $0.65$ ?

Calculator loads as you approach

Use Desmos if it helps you solve or check this problem.

Change a factor to a longer interval

Practice problem

An investment grows exponentially and is multiplied by $1.331$ every $9$ months. By what percent does the investment increase each year, rounded to the nearest tenth of a percent?

Your answer

Calculator loads as you approach

Work out the yearly percent in one expression, and round only at the end.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

The same difference each step means linear change. The same ratio means exponential change.
“Starts at $a$ and multiplies by $b$ every $k$ units” becomes $a\,b^{t/k}$ .
Turn a percent into a factor with $1\pm\frac{p}{100}$ . A factor below $1$ is the part that remains.
To change a factor to a new interval, use $b^{\text{new interval}/\text{old interval}}$ .
Once you have the new exponent, one expression can finish an awkward percent change, with no rounding until the end.
Use exact ratios for clean tables. Save exponential regression for messy real data where the ratios are close but not equal.

Related lesson

Later, for optional Desmos work with shifted exponential forms and regression, see Exponential models, transformations, and Log Mode.

Next lesson

Transform nonlinear functions

Connect translations, reflections, and stretches with equations and graphs.

Start next lesson

Practice

Practice this lesson

488 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

Start practice