Digital SAT · Nonlinear Graphs

How to Solve Exponential Growth and Decay on the Digital SAT

An exponential model looks like y equals a times b to the power x, where a is where you start and b is the factor you multiply by each period. Read the starting amount straight off, then turn the rate into b: add the rate for growth, subtract it for decay. Line the exponent up with the period the rate describes. When the choices are equations, graph your pick in Desmos and check its key points.

Written from Perfect1600’s analysis of every exponential growth and decay question in our bank·Method checked against the current Bluebook test
per test
2 per test
per test
typical difficulty
Medium to hard
typical difficulty
practice questions
46
practice questions
How we counted

Frequency reflects how often this question type appears on a full-length Digital SAT. The difficulty mix reflects every question of this type across our bank.

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What the question bank shows

a·bˣ
Start times a factor

a is where you begin; b is 1 plus the rate for growth, 1 minus for decay.

37%
Check on Desmos

About a third can be confirmed by graphing the model and reading its points.

the hazard
Read the rate right

Adding a percent instead of compounding it is the frequent slip.

How to recognize exponential growth and decay questions

  • A quantity grows or shrinks by a percentage or a fixed factor each period: interest, population, half-life, views.
  • The question asks which equation models it, or for a value after some periods.
  • Choices are equations of the form a times b to a power, differing in the base or the exponent.
  • Words like doubles, halves, increases by, or decays point to a factor, not a constant change.

Why students miss these

The pitfall is treating exponential change as linear and misreading the rate. An 8 percent yearly increase uses a factor of 1.08, not an added 8, and an 8 percent decay uses 0.92, not minus 8. The exponent gets mismatched to the time unit too, years where the rate is monthly. Since the choices differ only in the base or exponent, one wrong factor sinks an otherwise reasonable equation.

The step-by-step method

  1. 1

    Read the starting value

    The amount at time zero is a, the coefficient in front of the power.

  2. 2

    Turn the rate into a factor

    Growth is 1 plus the rate; decay is 1 minus the rate; doubling is 2, halving is one half.

  3. 3

    Match the exponent to the period

    The exponent counts rate-periods, so a monthly rate wants months, not years.

  4. 4

    Check a point or the graph

    Plug in a small time, or graph in Desmos, and confirm the starting value and one later point fit.

Solving it on Desmos

  1. Graph the candidate. Type the equation you suspect as y = ... to see its curve.
  2. Check the key points. Confirm the y-intercept is the starting value and a later point matches the description.
  3. Compare choices. Graph each option and keep the one whose start and growth fit; wrong factors miss the points.

Full Desmos walkthrough for exponential growth and decay

When to use it: Use Desmos to confirm which equation fits by checking key points, or to evaluate an amount after a given time. You build the model from the starting value and factor first; the graph confirms it.

Worked examples

Easy example

An investment of 500500 dollars grows by 3%3\% each year.

Which equation gives the value yy, in dollars, after xx years?
y=500(1.03)xy = 500(1.03)^{x}
B
y=500(0.03)xy = 500(0.03)^{x}
C
y=500(0.97)xy = 500(0.97)^{x}
D
y=500(1.3)xy = 500(1.3)^{x}

A: Correct. Growth of 3%3\% means a factor of 1+0.03=1.031 + 0.03 = 1.03, so y=500(1.03)xy = 500(1.03)^{x}.

B: Incorrect. This uses the rate 0.03 as the factor instead of 1+0.031 + 0.03.

C: Incorrect. This is a decay factor 10.031 - 0.03; the value grows, not shrinks.

D: Incorrect. This misplaces the decimal, using 1.31.3 (a 30%30\% increase) instead of 1.031.03.

Explanation

A 3%3\% yearly increase gives a growth factor of 1.031.03, so y=500(1.03)xy = 500(1.03)^{x}.

Medium example

A 60-milligram sample loses one-third of its mass each hour.

Which equation gives the mass mm, in milligrams, remaining after hh hours?
A
m=60(1.33)hm = 60(1.33)^{h}
B
m=60(13)hm = 60\left(\frac{1}{3}\right)^{h}
C
m=6013hm = 60 - \frac{1}{3}h
m=60(23)hm = 60\left(\frac{2}{3}\right)^{h}

A: Incorrect. A factor above 1 models growth, not a loss.

B: Incorrect. A factor of 13\frac{1}{3} removes two-thirds each hour, not one-third.

C: Incorrect. This is linear decay, not a one-third proportional loss.

D: Correct. Losing one-third leaves two-thirds, so the factor is 23\frac{2}{3}.

Explanation

Retaining 23\frac{2}{3} each hour: m=60(23)hm = 60\left(\frac{2}{3}\right)^{h}.

Hard example

A video has 500 views, and the number of views triples each day.

How many views will the video have after 3 days?
A
15001500
B
35003500
C
45004500
1350013500

A: Incorrect. This is the count after only 1 day: 500(3)500(3).

B: Incorrect. This treats growth as linear: 500+3000500 + 3000.

C: Incorrect. This is the count after only 2 days: 500(3)2500(3)^{2}.

D: Correct. 500(3)3=500(27)=13500500(3)^{3} = 500(27) = 13500.

Explanation

500(3)3=13500500(3)^{3} = 13500.

Detailed explanation

Tripling each day for 3 days multiplies by 33=273^{3} = 27, so 500×27=13,500500 \times 27 = 13{,}500. Stopping after 1 or 2 days, or adding instead of multiplying, gives the smaller values.

The common traps

PatternWhat it doesThe tell
Added instead of compoundedTreated a percentage change as a constant added each period.A percentage per period is a factor like 1.08, not repeated addition.
Wrong decay factorUsed 1 plus the rate for decay, or a negative base.Decay uses 1 minus the rate, a number between 0 and 1.
Mismatched time unitPut years in the exponent when the rate is monthly.The exponent counts the same periods the rate describes.
Start and factor swappedPlaced the rate where the starting amount belongs.At time zero the model must equal the starting amount.

Try it: two real questions

Question 1easy

A machine is worth 800800 dollars and loses 10%10\% of its value each year.

Which equation gives the machine's value yy, in dollars, after xx years?

Question 2easy

A culture begins with 12001200 bacteria, and the number triples every hour.

Which equation gives the number of bacteria yy after xx hours?

Question 3 is ready when you are

Keep going with more questions of this type, each with the same step-by-step reasoning. Free to start.

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Often confused with

Related reading

Common questions

How do you write an exponential growth or decay equation on the SAT?

Use y = a times b to the power x. Set a to the starting value, then b to 1 plus the rate for growth or 1 minus the rate for decay, with the exponent counting the rate's periods.

What is the growth or decay factor?

The number you multiply by each period. A 5 percent increase is 1.05; a 5 percent decrease is 0.95; doubling is 2 and halving is one half.

Can Desmos help with exponential models on the Digital SAT?

Yes. Graph the equation you think is right and confirm its starting value and a later point, or evaluate the amount after a given time. A fast check once you have the factor.

How do I know if a situation is exponential or linear?

If the quantity changes by a percentage or a fixed factor each period, it is exponential; if it changes by the same amount each period, it is linear. Doubles, halves, or percent per year signal exponential.

How do I handle a rate given per month when time is in years?

Keep the exponent in the rate's units. For monthly growth, count months; if the question asks in years, multiply the number of years by 12.

Practice exponential growth and decay the way it is tested

Start with the free 16-question diagnostic, then drill this type with step-by-step reasoning on every question.

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