Digital SAT Math · Advanced Math

Exponential functions

Digital SAT Math · Advanced Math

A linear model asks how much is added each period; an exponential model asks what is it multiplied by. That single swap — addition for multiplication — is the whole skill, and almost every wrong answer on it is a student applying the linear instinct to a multiplicative situation: adding a percent, dividing a factor, spreading a doubling evenly across the days it takes. Get the multiplier right and the rest is arithmetic, because the model is always the same three parts: a starting amount, a number it is multiplied by, and an exponent that counts how many times.

On the test


Domain	Advanced Math (score report)
What it looks like	A quantity that grows or shrinks by a percent, doubles, halves, or is tabulated at equal time steps — printed as a formula, a table, a curve, or a paragraph
Often asked	“Which function models…?”, “What is the best interpretation of…?”, “By what percent…?”, “Which equation could represent the relationship shown in the table?”, “How many … after $t$ …?”
Format	Multiple choice and student-produced response; the two live timed sets filed under this skill are mixed-format
Calculator	Genuinely useful here — powers of awkward decimals are the one place on this skill where Desmos saves real time, and the one place a hand slip is likely

Recognition cues: each year / every hour paired with a percent; doubles, triples, halves, half-life; decreases by 8% of its value; compounded annually; grows exponentially; a table whose entries multiply rather than step; a curve that bends but never turns around.

What this page owns

The College Board files quadratic and exponential behaviour together under one “nonlinear functions” heading, and most of the free web copies that heading onto a single page. We split them, because the two curves fail in opposite ways. A parabola turns around; an exponential never does — it flattens toward an invisible floor on one side and runs away on the other, which is why maximum and vertex never appear on this page and initial value and factor never leave it. When the answer is a turning point, that is quadratic functions. When the whole question is one percent change with no repetition, that is percentages. When a constant amount is added each period, that is linear functions. When the exponent contains the unknown and you are asked to solve for it, the Digital SAT keeps the numbers friendly enough to match bases or step through years — see the FAQ, and note that nothing on this test requires a logarithm.

Pattern recognition

Six shapes cover nearly everything:

Build from a percent — a story gives a starting amount and a percent per period; you write $y = a(1 \pm r)^t$ .
Build from a doubling or halving — a story gives a period (“every 5 days”, “half-life of 6 hours”); the period goes into the exponent’s denominator.
Read a printed model — the function is given and one number is wanted: the initial value, the factor, or an output at some time.
Interpret a printed model — nothing is computed; you say in words what the base or the coefficient means, in the units of the story.
Classify a table — equal differences or equal ratios, and then the rule that produced it.
Read a curve — the $y$ -intercept gives the initial value, the ratio of two marked outputs gives the base, and a third point checks you.

Linear or exponential? — the five-second decision. Ask one question: does the story add or multiply? Words that mean add — per, each additional, a constant rate of, increases by 40 units a year, a fixed fee → linear, $y = mx + b$ , constant first differences in a table, a straight line on a graph. Words that mean multiply — per year attached to a percent, doubles, halves, triples, decays, compounded, grows by a factor of → exponential, $y = ab^t$ , constant ratios in a table, a curve that steepens (or flattens) without turning. The tell in a table is mechanical, so use it rather than your eye: subtract consecutive outputs, and if those differences are not equal, divide consecutive outputs instead. One of the two will come out constant. And be careful about deciding from two rows — a line and an exponential can agree at two points and separate at the third, which is exactly how the best distractor on this skill is built.

Method

Name the three parts before writing anything. The starting amount $a$ , the per-period multiplier $b$ , and the unit of one step of the exponent. If you cannot say what one step of the exponent means in the story’s own words — one year, one month, one 6-hour period — you are not ready to write the model.
Turn the percent into a multiplier, never a base of its own. Growth of $r$ gives $b = 1 + r$ ; a loss of $r$ gives $b = 1 - r$ . A 7% rise is $1.07$ , a 7% fall is $0.93$ , and $0.07$ is neither.
Put the time unit in the exponent, not in the base. If the quantity multiplies by $k$ every $p$ units of time, the model is $a \cdot k^{t/p}$ . Dividing the factor by the period instead — treating “doubles every 5 days” as 40% a day — is wrong by a wide margin, because factors compound and amounts do not.
Anchor $a$ at $t = 0$ . The coefficient is the output when the exponent is zero, since $b^0 = 1$ . If the story hands you a value at some other time, you have a one-step equation to solve, not a coefficient to copy.
Answer the question that was asked. A factor (1.25), a rate (25%), and an amount (250 users) are three different answers, and all three will be offered. Check the units of your answer against the units in the question sentence.
Check by stepping. Exponential models are self-checking: multiply your way forward one period at a time and see whether you land on the value the model predicts. Two lines of arithmetic catch nearly every slip on this skill.

Words in the stem	Where it goes
initially, at the start, when first measured, in year 0	the coefficient $a$
increases by 6% each year	$b = 1.06$
decreases by 6% each year	$b = 0.94$
is 6% of its previous value	$b = 0.06$ — rare, and always a trap in disguise
doubles / triples every $p$ periods	$b = 2$ or $3$ , exponent $t/p$
half-life of $p$	$b = \tfrac12$ , exponent $t/p$
by a constant amount each period	not exponential at all — go to linear

Percent to multiplier, in both directions. Going in: up 3% → $1.03$ ; down 3% → $0.97$ ; up 150% → $2.5$ (the original plus one and a half of it); down 100% → $0$ . Coming back out is where items are actually won, because a multiplier has to be turned into a sentence: $1.4$ means up 40%, $0.4$ means down 60%, $0.96$ means down 4%, and $2$ means up 100%, not up 200%. Two habits keep this straight. First, any base below 1 is a decrease and any base above 1 is an increase, no matter how the sentence is phrased. Second, the percent in the sentence is the distance from 1, so read $|b - 1|$ and then decide the direction from which side of 1 the base sits on. Multipliers also compound rather than add: two consecutive 10% rises multiply by $1.1 \times 1.1 = 1.21$ , a 21% rise, and a 20% rise followed by a 20% fall gives $1.2 \times 0.8 = 0.96$ , a net 4% loss on the original.

The exponent is where the time unit lives. Three phrasings of the same model cause most of the unit damage. If a quantity multiplies by $k$ once per period and $t$ counts periods, the exponent is just $t$ . If $t$ counts something smaller than the period — days when the doubling takes 5 days — the exponent is $t/5$ , because only one fifth of a doubling happens per day. If $t$ counts something larger — years when the growth is quoted per month — the exponent is $12t$ . The test of a candidate model is always the same: substitute one period’s worth of $t$ and see whether you get exactly one multiplication by $k$ . And converting a rate between units is a power, not a product: 2% a month is $1.02^{12} \approx 1.268$ , about 26.8% a year, not 24%. Going the other way, an annual factor becomes a monthly one by taking a twelfth root, which is exactly the move that turns a two-year factor of $\tfrac14$ into an annual factor of $\tfrac12$ .

Two points give you the whole model. Given $f$ at two times, there is only one exponential through them. Subtract the exponents to find how many periods separate the points, divide the outputs to find the total factor across that gap, and take that root. From $f(2) = 12$ and $f(5) = 96$ : the gap is 3 periods, the total factor is $96/12 = 8$ , so $b^3 = 8$ and $b = 2$ . Then walk back to the initial value — $f(2) = ab^2 = 4a = 12$ , so $a = 3$ — and the model is $f(t) = 3 \cdot 2^t$ . Check it at the point you did not build it from: $3 \cdot 2^5 = 96$ ✓. The one thing never to do here is average: the value halfway in time between 12 and 96 is $f(3.5)$ , and it is $3 \cdot 2^{3.5} \approx 33.9$ , nowhere near the arithmetic mean of 54. Averages belong to lines.

Desmos on this skill — where it pays, and where it quietly wastes a minute. Two things it does better than you: raising an ugly decimal to a power (type 2000*1.05^9 and read $3102.66$ instead of multiplying nine times), and settling a crossing question, where plotting y=2000*1.05^x and y=2000+120x together shows you the year one passes the other far faster than a table of guesses. It is also the safest way to check a model you built — plot your candidate, then type the given points as $(x, y)$ pairs and see whether they land on the curve. Where it earns you nothing: an interpretation item, where the answer is a sentence and every choice plots identically or not at all; a build item, where the four choices are functions and the trap is a unit inside an exponent that Desmos will happily graph for you without comment; and any item whose numbers are clean powers, where stepping — 240, 120, 60, 30, 15 — is finished before the calculator has loaded. One habit worth keeping: when the answer is money, carry the exact value and round only at the end, because a two-cent rounding at year 8 is what turns a correct comparison into the wrong answer choice.

Worked example 1 — a percent becomes a multiplier

Stem. A used piano is bought for $3,200. Each year, its resale value is 12% less than its value the year before. (a) Write a function $V$ giving the resale value in dollars $t$ years after the purchase. (b) What is the value after 3 years, to the nearest cent?

Step 1 — name the three parts. Starting amount $a = 3200$ . One step of the exponent is one year. The multiplier is not $0.12$ : losing 12% means keeping the other 88%, so $b = 1 - 0.12 = 0.88$ .

Step 2 — write the model.

V(t) = 3200(0.88)^t

Step 3 — evaluate at $t = 3$ .

V(3) = 3200(0.88)^3 = 3200(0.681472) = 2180.7104

so about $2,180.71.

Check. Step year by year and watch the loss shrink: $3200 \to 2816 \to 2478.08 \to 2180.7104$ . The first year loses $384, the second $337.92, the third $297.37 — each is 12% of the current value, which is what the sentence said, and the three losses are not equal, which is what makes this exponential rather than linear.

Trap watch. $3200(0.12)^t$ keeps 12% a year and reaches $5.53 in three years. $3200(1.12)^t$ grows the piano’s value. And $3200 - 384t$ — a fixed $384 a year — agrees exactly at $t = 1$ and is wrong everywhere after, which is why checking a model at one point proves nothing.

Worked example 2 — same two rows, different futures

Stem. Two quantities are recorded at four equally spaced times:

$x$	0	1	2	3
$f(x)$	16	24	36	54
$g(x)$	16	24	32	40

One is linear and the other is exponential. Which is which, and what are their rules?

Step 1 — first differences. For $g$ : $24-16 = 8$ , $32-24 = 8$ , $40-32 = 8$ . Equal differences, so $g$ is linear with slope 8 and intercept 16: $g(x) = 16 + 8x$ .

Step 2 — for $f$ the differences fail, so try ratios. $24-16 = 8$ , $36-24 = 12$ , $54-36 = 18$ — not constant. Divide instead: $24/16 = 1.5$ , $36/24 = 1.5$ , $54/36 = 1.5$ . Equal ratios, so $f$ is exponential with base $1.5$ , and its value at $x=0$ is the coefficient:

f(x) = 16(1.5)^x

Check. Test each rule at a row you did not use to build it: $f(3) = 16(1.5)^3 = 16(3.375) = 54$ ✓ and $g(2) = 16 + 16 = 32$ ✓. Note what the first two columns do — the two quantities are identical at $x = 0$ and $x = 1$ , and only separate at $x = 2$ . Base $1.5$ also means $f$ grows 50% per step, which is a sentence the item may ask for instead of the rule.

Trap watch. Deciding from two rows is the whole trap here, and it is the trap in the practice set too. The other classic is calling $f$ linear “because the differences go up steadily” — increasing differences are the signature of a nonlinear function, not of a line, and the differences of an exponential are themselves exponential (8, 12, 18 has the same ratio 1.5).

Worked example 3 — a doubling period lands in the exponent

Stem. The mass of algae in a water tank doubles every 3 days. There are 200 grams now. (a) Write a function $m$ giving the mass in grams $d$ days from now. (b) What is the mass 12 days from now?

Step 1 — the factor and the period are two different numbers. The factor is $2$ . The period is $3$ days. The variable $d$ counts days, which are smaller than a period, so the exponent must count periods: in $d$ days there are $d/3$ of them.

Step 2 — write the model.

m(d) = 200 \cdot 2^{d/3}

Step 3 — evaluate at $d = 12$ . Twelve days is $12/3 = 4$ doubling periods:

m(12) = 200 \cdot 2^{4} = 200(16) = 3200 \text{ grams}

Check. Step in 3-day jumps: $200 \to 400 \to 800 \to 1600 \to 3200$ , four arrows for the four periods in 12 days ✓. Test the model on the sentence that produced it: $m(3) = 200 \cdot 2^{1} = 400$ , a single doubling in 3 days, exactly as stated.

Trap watch. $200 \cdot 2^{3d}$ doubles three times a day and reaches 200 · 2³⁶ in 12 days. $200 \cdot 2^{d}$ ignores the period and gives 819,200. $200 \cdot 2 \cdot 4 = 1600$ treats four doublings as four multiplications-by-two added up rather than compounded — the linear instinct again, one period short of the truth.

Practice

Answer before opening the explanation. Two items are student-produced response, matching the mixed format of the live timed sets on this skill; three carry a figure — a table and two curves — because on this skill the display is often the question rather than an illustration. Every wrong choice below is one named error: a percent used as a base, a factor divided instead of an exponent, an initial value read one step late, a rate quoted for the wrong unit of time. Log the error, not the item number.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

A proofreader estimates that the number of typos still in a manuscript after p proofreading passes is T(p) = 384(0.75)ᵖ. According to this model, how many typos were in the manuscript before any proofreading passes were made?

Show the answer Choice B

Why it is right

Before any passes have been made, p = 0. Any nonzero number raised to the power 0 equals 1, so T(0) = 384(0.75)⁰ = 384(1) = 384 typos. This is why the number multiplying the power is called the initial value: in a model written a·bᵖ, the constant a is the output at p = 0, and the base b only starts to act once the exponent is at least 1.

Why each other choice fails

Choice A: This is T(1) = 384(0.75) = 288, the count after the first pass has already removed a quarter of the typos. The question asks for the count before any pass, which is the exponent p = 0, not p = 1.
Choice C: This divides by 0.75 instead of leaving the initial value alone, as though the manuscript had already been proofread once when the model started. The model is written from the unproofread manuscript, so no step backward is needed: 384 already is the starting count.
Choice D: This is 0.25 × 384 = 96, the number of typos the first pass removes rather than the number present at the start. The factor 0.75 says three quarters survive each pass; the quarter that disappears is a change, not a starting amount.

Question 2 Standard

A tablet is bought for $480. Each year after the purchase, its resale value is 15% less than its resale value the year before. Which function models the resale value V(t), in dollars, t years after the purchase?

Show the answer Choice A

Why it is right

Losing 15% of the current value means keeping the other 85% of it, so each year's value is 0.85 times the previous year's value. Repeating that multiplication t times gives V(t) = 480(0.85)ᵗ. Check the first two years against the words: 480(0.85) = 408, which is 72 less than 480 and therefore exactly 15% of 480; then 408(0.85) = 346.80, which is 61.20 less, exactly 15% of 408 rather than of the original price.

Why each other choice fails

Choice B: A base greater than 1 makes the value grow, and 1.15 grows it by 15% a year: V(1) = 552, more than the purchase price. The 1 + r form is for an increase; a decrease of 15% needs 1 − 0.15 = 0.85.
Choice C: Using the percent itself as the base keeps only 15% of the value each year, a 85% loss: V(1) = 72. The rate and the multiplier are different numbers, and the multiplier is what the base of an exponential model must be.
Choice D: This subtracts a fixed $72 every year, which is 15% of the original price rather than 15% of the current value. It agrees with the correct model at t = 1 (both give 408) and then drifts: at t = 2 it gives 336 instead of 346.80, and it reaches $0 in under seven years, while a percent loss never quite reaches zero.

Question 3 Standard

A technician starts a yeast culture containing 60 cells. The number of cells triples every hour. How many cells are in the culture 4 hours after it is started?

Show the answer Choice C

Why it is right

Tripling every hour means multiplying by 3 once per hour, so after 4 hours the starting count has been multiplied by 3 four times: 60 · 3⁴ = 60 · 81 = 4,860 cells. Counting the hours one at a time gives the same thing and is worth doing once as a check: 60 → 180 → 540 → 1,620 → 4,860, which is four arrows for four hours.

Why each other choice fails

Choice A: This triples only once, giving the count after 1 hour rather than after 4. The exponent counts how many times the multiplication happens, so the 4 hours have to appear as a power, not be ignored.
Choice B: This computes 60 · 3 · 4, treating the growth as a constant rate of tripling per hour the way a linear model treats a constant amount per hour. Repeated multiplication is a power, 3⁴ = 81, not a product 3 × 4 = 12.
Choice D: This is 60 · 3³, the count after 3 hours. The step from 3 hours to 4 hours multiplies by 3 one more time, and 1,620 · 3 = 4,860 recovers the correct answer, so the slip is one hour of counting.

Question 4 Standard

The table gives the number of users of two apps, A and B, at the end of each of the first three weeks after launch. The number of users of app B grows exponentially. Which equation gives the number of users y of app B, w weeks after launch?

Weeks after launch, w	App A users	App B users
0	256	256
1	320	320
2	384	400
3	448	500

Figure 1. Users of two apps over the first three weeks. The two columns agree at weeks 0 and 1 and separate afterward.

Show the answer Choice D

Why it is right

Divide each app B entry by the one before it: 320/256 = 1.25, 400/320 = 1.25, and 500/400 = 1.25. A constant ratio is what exponential growth means, so the base is 1.25, and the value at w = 0 is 256, so y = 256(1.25)ʷ. Test it away from the row you used: at w = 3 it gives 256(1.25)³ = 256(1.953125) = 500, matching the table.

Why each other choice fails

Choice A: This is app A's rule, not app B's: app A adds a constant 64 users a week, which is why its differences 64, 64, 64 are equal while its ratios are not. It matches app B for the first two rows and then falls short, giving 384 and 448 where app B has 400 and 500.
Choice B: This uses the 25% in the growth factor as the base itself. A base below 1 shrinks the quantity — it predicts 64 users after one week — while the table shows the count rising. The base of a growth model is 1 + r, not r.
Choice C: This swaps the initial value with the base. It happens to give 320 at w = 1, which is why it survives a single check, but at w = 0 it gives 1.25 users instead of 256, and at w = 2 it gives 81,920. The number being multiplied repeatedly is the base, and here that number is the ratio 1.25.

Question 5 Standard

A farm cooperative bought a used tractor. The tractor's value V, in dollars, t years after the purchase is modeled by V(t) = 24000(0.94)ᵗ. Which of the following is the best interpretation of 0.94 in this model?

Show the answer Choice B

Why it is right

The base of an exponential model is the number the quantity is multiplied by once per unit of time. Multiplying by 0.94 keeps 94% of the value, which is the same as losing the remaining 6%. So the tractor holds 94% of its value each year: 24,000 becomes 22,560, which becomes 21,206.40, each figure being 0.94 times the one before rather than a fixed dollar amount below it.

Why each other choice fails

Choice A: This reads the multiplier as if it were the rate. Losing 94% a year would leave only 6% behind, a multiplier of 0.06 and a value of $1,440 after one year — a collapse the model does not describe.
Choice C: This applies 6% of the original $24,000 every year, which is a straight-line loss of a fixed amount. It is right for the first year only: the model's second-year loss is 6% of 22,560, or $1,353.60, and every later loss is smaller still.
Choice D: The direction is wrong. A base below 1 shrinks the quantity, so 0.94 is a decrease; the 6% in this choice is correctly computed as 1 − 0.94 but then attached to growth instead of decay.

Question 6 Harder

A survey counts 90 beetles in a wheat field. The number of beetles in the field doubles every 5 days. Which function models the number of beetles N(d) in the field d days after the survey?

Show the answer Choice A

Why it is right

The exponent must count doubling periods, not days. In d days there are d/5 periods of 5 days, so the count is 90(2)^(d/5). Test it on the sentence it came from: at d = 5 the exponent is 1 and the model gives 180, one doubling; at d = 10 the exponent is 2 and it gives 360, two doublings. Dividing by the length of the period is the general move — a quantity that multiplies by k every p units of time is modeled by a·k^(t/p).

Why each other choice fails

Choice B: Multiplying the exponent by 5 instead of dividing gives a doubling every fifth of a day. At d = 5 it predicts 90(2)²⁵, about 3 billion beetles, when the field should hold 180.
Choice C: This doubles the beetles every day, ignoring the 5 entirely. It reaches 2,880 after 5 days instead of 180, so it is off by a factor of 16 after less than a week.
Choice D: This converts the doubling to a daily rate by dividing the factor rather than the exponent, as though 2 over 5 days meant about 40% a day. Compounding 1.4 for five days gives 1.4⁵ ≈ 5.38, more than five times the population instead of twice; a factor cannot be spread across a period by division.

Question 7 Harder

The graph of y = f(x) is shown in the xy-plane, where f is an exponential function. Which equation defines f?

Figure 2. The graph of y = f(x), with three points marked.

Show the answer Choice C

Why it is right

Read the initial value off the y-intercept: the curve crosses the y-axis at (0, 5), so a = 5. Then read the base from a marked point one unit to the right: f(1) = 10, and 10/5 = 2, so b = 2, giving f(x) = 5(2)ˣ. Confirm with the marked point the base was not built from: f(3) = 5(2)³ = 5(8) = 40, which is exactly the third marked point.

Why each other choice fails

Choice A: This swaps the initial value and the base. Swapping them always leaves the value at x = 1 unchanged, since 2 · 5 and 5 · 2 are the same, so this choice does pass through (1, 10) — but its y-intercept is 2 instead of 5, and it reaches 250 at x = 3 where the graph shows 40.
Choice B: This is the line through the first two marked points, which is why it survives both of them: it gives 5 at x = 0 and 10 at x = 1. It fails at the third: a line adds 5 each step and reaches 20 at x = 3, while the curve doubles each step and reaches 40. Two points never distinguish a line from an exponential; the third does.
Choice D: This reads the initial value from the point (1, 10) instead of the y-intercept. The initial value is the output at x = 0, and this function gives 10 there, twice what the graph shows.

Question 8 Harder

The number of registered electric vehicles in a city is modeled by E(m) = 900(1.02)ᵐ, where m is the number of months after January 2024. Which of the following is the best interpretation of this model?

Show the answer Choice D

Why it is right

Two readings settle it. The coefficient 900 is the output at m = 0, which the model defines as January 2024. The base 1.02 is 1 + 0.02, so each step of the exponent multiplies the count by 1.02, an increase of 2% — and one step of m is one month, because the stem says m counts months. Growth of 2% per month is therefore the whole of what the base says.

Why each other choice fails

Choice A: This reads the increase as an added amount instead of a multiplier. Adding 2 vehicles a month would be E(m) = 900 + 2m, a straight line; the model multiplies, so the monthly increase starts at 18 vehicles and grows.
Choice B: The percent is right and the time unit is wrong. Growth of 2% per year would put the year in the exponent, as 900(1.02)^(m/12); as written, the exponent advances a full step every month.
Choice C: This converts months to years by multiplying the monthly percent by 12, which is how a constant amount converts, not a constant factor. Twelve months of 2% growth multiply the count by 1.02¹² ≈ 1.268, an annual increase of about 26.8%, not 24%.

Question 9 Harder Student-produced response

A patient is given a 240-milligram dose of a medication. The amount of the medication remaining in the patient's bloodstream is halved every 6 hours. How many milligrams remain 24 hours after the dose is given?

Your answer

Show the answer 15

Why it is right

In 24 hours there are 24/6 = 4 half-life periods, so the starting amount is halved four times: 240 · (1/2)⁴ = 240/16 = 15 milligrams. Stepping through the periods confirms it and is the safer route under time pressure: 240 → 120 at 6 hours → 60 at 12 hours → 30 at 18 hours → 15 at 24 hours, four halvings for four periods.

Answers students type instead

30: Stops after three halvings, the amount left at 18 hours. One more 6-hour period fits inside 24 hours, and halving 30 gives the correct 15.
60: Divides 240 by 4, the number of periods, instead of halving four times. Repeated halving is division by 2⁴ = 16, not by 4; dividing by the count of periods would describe a quantity being shared out, not decaying.
120: Halves once and stops, answering for 6 hours rather than 24. The exponent counts periods, so the 24 hours in the question has to be converted into the four halvings it contains.

Question 10 Hardest

The graph shows the mass A, in grams, of a sample of a radioactive material remaining t years after the first measurement, where A is an exponential function of t. According to this model, by what percent does the mass of the sample decrease each year?

Figure 3. Mass of the sample remaining over six years, with three points marked.

Show the answer Choice B

Why it is right

The marked points give the two-year factor: from (0, 640) to (2, 160) the mass is multiplied by 160/640 = 1/4. Two years of the annual factor b produce that, so b² = 1/4 and b = 1/2. A yearly multiplier of 0.5 keeps half and therefore loses half, a 50% decrease each year. The third marked point checks it: two more years from 160 at the same rate give 160(1/2)² = 40, exactly the point at (4, 40).

Why each other choice fails

Choice A: This reports the fraction of mass that survives two years, 160/640 = 0.25, as if it were the percent lost in one year. A retained fraction and a percent decrease are complements, and neither of them is annual here.
Choice C: This takes the two-year decrease of 75% and splits it evenly across the two years. Percent changes compound rather than add: two years at 37.5% would leave 0.625² = 0.390625 of the mass, about 250 grams at t = 2, not 160.
Choice D: This is the correct percent for the wrong length of time. The mass does fall 75% between t = 0 and t = 2, but the question asks per year, and the answer must be the single-year factor, which is the square root of the two-year one.

Question 11 Hardest Student-produced response

The number of trees infected by a blight in a forest is modeled by an exponential function of the number of years since 2021. There were 7 infected trees in 2021 and 189 infected trees in 2024. According to the model, how many trees were infected in 2022?

Your answer

Show the answer 21

Why it is right

Write the model as N(t) = 7bᵗ with t years after 2021, since 7 is the count at t = 0. The year 2024 is t = 3, so 7b³ = 189, giving b³ = 27 and b = 3. Then 2022 is t = 1 and N(1) = 7(3) = 21. Check the whole chain forward: 7 → 21 → 63 → 189, three multiplications by 3 across the three years the stem describes.

Answers students type instead

3: This is the growth factor b, not a number of trees. The factor answers 'how many times as many each year', while the question asks for a count, which needs the factor applied to the 2021 total.
27: This is the total factor across the three years, 189/7 = 27. It tells you the population multiplied by 27 between 2021 and 2024, so it is the cube of the annual factor, not the annual factor and not a count of trees.
63: This is the model's count for 2023, one year too late. It comes from stepping back once from 189 instead of forward once from 7; stepping back a second time reaches 21, the year the question asks about.

Question 12 Hardest

On the same day, Ravi puts $2,000 into an account that earns 5% interest per year, compounded annually, and Nina puts $2,000 into an account that grows by exactly $120 each year. Neither makes any other deposit or withdrawal. After how many whole years is the balance in Ravi's account first greater than the balance in Nina's account?

Show the answer Choice C

Why it is right

Ravi's balance is 2000(1.05)ⁿ and Nina's is 2000 + 120n. Nina starts ahead because 5% of $2,000 is only $100 against her $120, and she stays ahead while Ravi's balance is small. At n = 8, Ravi has 2000(1.05)⁸ = $2,954.91 against Nina's $2,960 — still behind by about five dollars. At n = 9, Ravi has 2000(1.05)⁹ = $3,102.66 against Nina's $3,080, so year 9 is the first year he is ahead. This is the general shape: a percent model eventually passes any fixed-amount model, but 'eventually' can be most of a decade.

Why each other choice fails

Choice A: Assumes the percent model leads from the start. It does not: after one year Ravi has $2,100 and Nina has $2,120, because 5% of the opening balance is $100, less than Nina's fixed $120.
Choice B: This is the year Ravi's yearly interest first passes $120: his balance reaches $2,431.01, so the next year's interest is $121.55. Growing faster is not the same as being ahead — Nina is still $49 in front at that point, and Ravi needs five more years to erase the lead he built up while behind.
Choice D: A near miss that comes from rounding the eighth-year balance. 1.05⁸ = 1.477455…, so Ravi has $2,954.91 against $2,960 and is still $5.09 behind. Rounding 1.05⁸ to 1.48 makes his balance come out as exactly $2,960 — a tie rather than a lead — and a tie read as a win costs the point.

Common mistakes

Using the percent as the base — 8% growth is $1.08$ and 8% decay is $0.92$ ; $0.08$ is a 92% collapse. The base is a multiplier, and a multiplier near 1 means a small change.
Reading a multiplier back as the wrong percent — $0.85$ is a 15% decrease, not an 85% one; $1.6$ is a 60% increase, not 160%. The percent is the distance from 1.
Adding a percent instead of multiplying — subtracting a fixed amount computed from the original value turns a decay model into a line. It agrees for exactly one period and then drifts, which is why it is so easy to miss.
Dividing the factor by the period — “doubles every 5 days” is not 40% a day. Periods divide the exponent: $2^{d/5}$ .
Converting a rate between time units by multiplying — 2% a month is not 24% a year, it is $1.02^{12} - 1 \approx 26.8\%$ . Rates compound.
Reading the initial value from the wrong place — $a$ is the output when the exponent is zero. On a graph that is the $y$ -intercept, not the first marked point; in a table it is the row where the input is 0, not the first row.
Deciding linear versus exponential from two rows — two points fit both. Check a third, and check ratios once the differences have failed.
Answering with the factor when the question wants a count (or the reverse) — $b = 3$ , “3 times as many”, and “21 trees” are three different answers to three different questions.
Averaging two outputs to get the one between them — the middle of an exponential is a geometric mean, not an arithmetic one. Between 12 and 108 sits 36, not 60.
Believing exponential always beats linear immediately — it always wins eventually and can lose for years first. Test the specific year the item asks about.
Rounding a compounded value too early — a two-cent rounding at year 8 flips a comparison. Keep the exact power until the last line.
Expecting a turning point — an exponential has no maximum, no minimum and no vertex; if the curve you are reading turns around, you are on the quadratic page’s skill, not this one.

FAQ

How do I tell exponential from linear in one look? Look at what the story does per period: adds a fixed amount (linear) or multiplies by a fixed number (exponential). In a table, subtract first — if the differences are equal it is linear — and divide only when the differences fail. On a graph, a line has a constant steepness and an exponential’s steepness changes without the curve ever turning around.

Do I need logarithms? No. The Digital SAT never requires one. When an exponent has to be found, the numbers are built to be matched ( $2^x = 32$ gives $x=5$ ) or stepped through (multiply year by year until you pass the target), and Desmos will show you the crossing point if the stepping gets long.

What is the difference between the growth rate and the growth factor? The rate is the percent change, $r$ ; the factor is what you multiply by, $b = 1 + r$ . An item that asks “by what percent” wants $r$ , an item that asks “which function” wants $b$ , and the most common way to lose the point is to write the right number in the wrong role.

Where does the initial value show up on a graph? At the $y$ -intercept, because that is the output when the exponent is 0. It is worth naming explicitly, since an exponential curve gets very close to the horizontal axis on one side, and it is easy to read the first marked point instead of the actual intercept.

“Compounded annually” — is that a different formula? Not for our purposes. Interest compounded once a year is exactly $a(1+r)^t$ . The general $A = P(1 + r/n)^{nt}$ form is a business-math staple that the current test does not require; what it does ask is whether you can compare a compounding account with a fixed-dollar one, which is a stepping problem.

Can an exponential model ever reach zero? No — a positive quantity multiplied by a positive factor stays positive forever, which is why a decay model flattens toward the axis but never touches it. A model that hits zero at a specific time is linear, and noticing that difference is enough to eliminate a choice on sight.

Two points, three unknowns — how is the model determined? There are only two unknowns in $y = ab^t$ , so two points are exactly enough. Divide the outputs to get the factor across the gap, take the root matching the number of periods, then back-substitute for $a$ . See the box above for a worked instance.

Grid-in format? Enter the exact value; fractions are accepted as fractions. The most common grid-in loss on this skill is entering the growth factor where a count was wanted, and the second most common is stopping one period short — with no answer choices, nothing warns you that you answered for the wrong year.

Coming next in this domain: interpreting nonlinear graphs.

Editorial team · last reviewed 2026-08-06

Exponential functions

On the test

What this page owns

Pattern recognition

Method

Worked example 1 — a percent becomes a multiplier

Worked example 2 — same two rows, different futures

Worked example 3 — a doubling period lands in the exponent

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

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Question 3 Standard

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Question 4 Standard

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Question 5 Standard

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Question 6 Harder

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Question 7 Harder

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Question 8 Harder

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Question 9 Harder Student-produced response

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Answers students type instead

Question 10 Hardest

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Question 11 Hardest Student-produced response

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Question 12 Hardest

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Common mistakes

FAQ

Related skills