How do I tell if a table is exponential or linear?

Check consecutive values. If they share a common difference (you add the same amount each step), it's linear. If they share a common ratio (you multiply by the same factor each step), it's exponential. For example 4, 8, 16, 32 doubles each time (ratio 2, exponential), while 4, 8, 12, 16 adds 4 each time (linear).

What's the difference between the base and the rate?

The base b is the multiplier per period; the rate r is the percent change. They relate by b = 1 + r for growth or b = 1 - r for decay. A base of 1.07 is a 7% growth rate; a base of 0.93 is a 7% decay rate. Answer choices often list the base as a decoy for the rate, so convert carefully.

Why do I raise the base to a power to change time units?

Because the exponent rule b^mt = (b^m)^t lets you regroup how many times the factor is applied. If something multiplies by 1.01 every month, then over a year it multiplies by 1.01 twelve times, which is (1.01)^12 — the effective annual factor. This is why compounding makes the yearly rate larger than 12 times the monthly rate.

Can I just use the Desmos calculator on these questions?

Often yes — you can graph a candidate model to confirm it passes through given points, or evaluate expressions like (1.006)^12 directly. But for interpretation questions and equivalent-expression questions, recognizing the structure of y = a· b^t is faster and less error-prone than testing every choice.

DSAT-2.4

Exponential Functions, Growth & Decay

Master Digital SAT exponential models y = a·bᵗ: write growth and decay equations, read rates from formulas or tables, and rewrite compounded exponents to find effective per-period rates.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Exponential Functions, Growth & Decay, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Exponential Functions, Growth & Decay free

What this lesson covers

Exponential functions describe anything that multiplies by a fixed factor over equal time steps — populations, investments, radioactive decay, and depreciating cars. On the Digital SAT, these questions reward students who can move fluidly between a real-world story, an equation like

y = a\cdot b^{t}

, and a table of values.

This lesson shows you how to build the model from scratch, pull the growth or decay rate straight out of the base, and handle the trickiest version the test loves: rewriting an exponent so a monthly rate becomes an annual rate (or vice versa). Get comfortable with these moves and you'll turn slow word problems into quick, reliable points.

The Core Model: y = a·bᵗ

Every exponential function on the SAT fits the form

y = a\cdot b^{t}

, where

a

is the initial value (the amount when

t = 0

) and

b

is the growth/decay factor (the number you multiply by each time period).

The single most important rule: plug in

t = 0

and the exponent makes

b^{0} = 1

, so

y = a

. That means

a

is always the starting amount — the value before any time has passed. The test constantly asks you to interpret

a

in context, and it is simply the y-intercept of the curve.

The base

b

controls direction. If

b > 1

, the quantity grows. If

0 < b < 1

, the quantity decays. The base can never be negative or zero in these models.

Feature	Growth	Decay
Base $b$	$b > 1$	$0 < b < 1$
Example	$y = 200(1.05)^{t}$	$y = 200(0.95)^{t}$
Behavior	Increases each period	Decreases each period

Contrast this with a linear model,

y = mx + b

, which adds a constant each step. Exponential models multiply by a constant each step. The SAT often shows a table and asks whether the pattern is linear or exponential — check whether successive values share a common difference (linear) or a common ratio (exponential).

Reading the Rate from the Base

The base

b

hides a percent rate. Write

b = 1 + r

for growth or

b = 1 - r

for decay, where

r

is the decimal rate per period.

For growth, a base of

1.08

means

r = 0.08

, an 8% increase each period. For decay, a base of

0.86

means

1 - r = 0.86

, so

r = 0.14

, a 14% decrease each period. A common misconception is reading

0.86

as an 86% decrease — it actually means 86% remains, so only 14% is lost.

So the full context-rich form is:

y = a\,(1 \pm r)^{t}

Example builds you should be able to do instantly: a population of 500 growing 3% per year becomes

y = 500(1.03)^{t}

. A 500 mg dose decaying 20% per hour becomes

y = 500(0.80)^{t}

.

When a question gives a table, find the common ratio by dividing any value by the previous one. If a bacteria count goes 40, 60, 90, 135, each term is

1.5

times the last, so

b = 1.5

and the growth rate is 50% per step. The initial value

a

is the value at

t = 0

— if the table starts at

t = 1

, work backward by dividing by

b

to recover

a

Rewriting Compounded Exponents

The hardest SAT exponential questions change the time unit. You know a rate per year but need it per month, or the model uses

t

in months but the question asks about yearly change.

The key algebra tool is the exponent rule

b^{mt} = (b^{m})^{t}

. If a model is written with time in one unit, you can regroup the exponent to reveal the factor for a different unit.

Suppose

y = 800(1.02)^{12t}

where

t

is in years, so the exponent counts 12 months per year. Rewrite it as

y = 800\big((1.02)^{12}\big)^{t}

. Now

(1.02)^{12} \approx 1.268

, meaning the effective annual growth is about 26.8%, even though the monthly rate is only 2%.

Going the other direction: given an annual model

y = a(1.268)^{t}

, the monthly factor is

1.268^{1/12} \approx 1.02

Task	What to do
Monthly base to annual	Raise base to the 12th power
Annual base to monthly	Raise base to the $1/12$ power
General: per $n$ sub-periods	Use $(b)^{n}$ or $(b)^{1/n}$

Watch how the exponent is structured. If

t

represents years but appears as

\frac{t}{12}

, then time is being divided, so each unit of

t

triggers a fractional power. Match the exponent's coefficient to the number of periods per unit.

How the Digital SAT Tests This

Expect three flavors of questions. First, interpretation: given

y = 1200(0.91)^{t}

, identify what 1200 or 0.91 means (starting value; 9% decrease per period). Second, model-building: translate a word problem or table into an equation and pick the matching answer choice. Third, rate conversion: rewrite an expression to find an equivalent form with a different time base, often phrased as "which expression shows the monthly growth rate?"

Strategy tips that save time: always test

t = 0

to confirm the initial value matches. When choices differ only by base, compute the common ratio from two data points. For "equivalent expression" problems, use

b^{mt} = (b^{m})^{t}

rather than plugging numbers.

A frequent trap is confusing the factor with the rate. The answer choice

0.91

is the factor; the rate is

9\%

. Another trap: the SAT may offer both

500(1.05)^{t}

and

500(1.5)^{t}

— a 5% rate gives

1.05

, not

1.5

. Read the decimal carefully.

The calculator (Desmos, built into the test) is powerful here: you can graph a proposed model and check that it passes through given points, or evaluate

(1.02)^{12}

directly. But recognizing the structure is faster than guess-and-check on most items.

Key terms

Initial value ( $a$ ).: The quantity when $t = 0$ ; the y-intercept of the exponential curve, since $b^{0} = 1$ .
Growth/decay factor ( $b$ ).: The constant multiplier applied each period. Growth when $b > 1$ , decay when $0 < b < 1$ .
Growth rate ( $r$ ).: The decimal percent increase per period, where $b = 1 + r$ .
Decay rate ( $r$ ).: The decimal percent decrease per period, where $b = 1 - r$ ; a base of $0.85$ means $r = 0.15$ .
Common ratio.: The constant factor between consecutive terms in a table; found by dividing any value by the previous one.
Effective per-period rate.: The overall rate for a chosen time unit, found by rewriting the exponent, e.g. annual factor $= (\text{monthly factor})^{12}$ .
Exponential vs. linear.: Exponential models multiply by a constant each step (common ratio); linear models add a constant each step (common difference).

Worked example

A savings account earns interest so that its balance is modeled by

B = 2500(1.006)^{12t}

, where

t

is measured in years. Which expression gives the balance as an equivalent function showing the effective annual growth factor, and what is the approximate annual interest rate?

Start by recognizing the exponent structure. Here

t

is in years, but the base

1.006

is applied

12t

times — the 12 tells us interest compounds monthly, at 0.6% per month.

To find the effective annual factor, group the exponent using

b^{12t} = (b^{12})^{t}

B = 2500\big((1.006)^{12}\big)^{t}

Now compute the inner value:

(1.006)^{12}

. Using the calculator,

(1.006)^{12} \approx 1.0744

.

So the equivalent annual model is

B \approx 2500(1.0744)^{t}

.

Interpret the base with

b = 1 + r

1.0744 = 1 + r

, so

r \approx 0.0744

, meaning about a 7.44% effective annual interest rate.

Notice this is more than

12 \times 0.6\% = 7.2\%

— that gap is the effect of compounding. The SAT wants you to see that raising to the 12th power, not just multiplying by 12, gives the true annual growth. The initial deposit

a = 2500

is unchanged because rewriting the exponent never touches the leading coefficient.

Practice questions

A car purchased for 24,000 dollars loses 15% of its value each year. Which equation models its value

V

after

t

years?

$V = 24000(1.15)^{t}$
$V = 24000(0.15)^{t}$
$V = 24000(0.85)^{t}$
$V = 24000(15)^{t}$

Answer: $V = 24000(0.85)^{t}$

This is decay, so the base must be between 0 and 1. Losing 15% means 85% remains each year, giving a factor of

1 - 0.15 = 0.85

. The choice

1.15

describes growth, and

0.15

would mean 85% is lost each year (far too fast). The initial value 24,000 stays as the coefficient

a

The number of members in a club is given in the table below, where

t

is years since founding.

$t$	Members
0	80
1	120
2	180
3	270

Write an exponential model for the number of members, and state the annual growth rate as a percent.

Answer: $y = 80(1.5)^{t}$ , a 50% annual growth rate.

Find the common ratio by dividing consecutive terms:

120/80 = 1.5

180/120 = 1.5

270/180 = 1.5

. The constant ratio confirms exponential growth with base

b = 1.5

. The initial value at

t = 0

is 80, so

y = 80(1.5)^{t}

. Since

b = 1 + r

, we get

r = 0.5

, or 50% growth each year.

A medication's concentration is modeled by

C = 60(0.5)^{t/6}

, where

t

is in hours. What does the value 6 in the exponent represent?

The initial concentration in milligrams
The number of hours for the concentration to halve
The percent lost each hour
The number of doses taken

Answer: The number of hours for the concentration to halve

Rewrite as

C = 60\big((0.5)^{1/6}\big)^{t}

to see the hourly factor, but the cleaner reading is: when

t = 6

, the exponent is

6/6 = 1

, giving

C = 60(0.5) = 30

— exactly half of 60. So every 6 hours the amount is multiplied by

0.5

, meaning 6 is the half-life. The initial concentration is 60, not 6.

FAQ

How do I tell if a table is exponential or linear?: Check consecutive values. If they share a common difference (you add the same amount each step), it's linear. If they share a common ratio (you multiply by the same factor each step), it's exponential. For example 4, 8, 16, 32 doubles each time (ratio 2, exponential), while 4, 8, 12, 16 adds 4 each time (linear).
What's the difference between the base and the rate?: The base $b$ is the multiplier per period; the rate $r$ is the percent change. They relate by $b = 1 + r$ for growth or $b = 1 - r$ for decay. A base of $1.07$ is a 7% growth rate; a base of $0.93$ is a 7% decay rate. Answer choices often list the base as a decoy for the rate, so convert carefully.
Why do I raise the base to a power to change time units?: Because the exponent rule $b^{mt} = (b^{m})^{t}$ lets you regroup how many times the factor is applied. If something multiplies by $1.01$ every month, then over a year it multiplies by $1.01$ twelve times, which is $(1.01)^{12}$ — the effective annual factor. This is why compounding makes the yearly rate larger than 12 times the monthly rate.
Can I just use the Desmos calculator on these questions?: Often yes — you can graph a candidate model to confirm it passes through given points, or evaluate expressions like $(1.006)^{12}$ directly. But for interpretation questions and equivalent-expression questions, recognizing the structure of $y = a\cdot b^{t}$ is faster and less error-prone than testing every choice.

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