SAT® Exponential Growth and Decay: Rates, Time Units, and Model Meaning
SAT Clarity Editorial ·
Editorial calendar: 2026-06-18. Actual publication date is shown above.
An exponential model changes by the same multiplicative factor over equal time intervals. A linear model changes by the same additive amount. That distinction explains why a 6% annual increase uses a factor of 1.06, why a 6% annual decrease uses 0.94, and why changing years to months changes the exponent or the base. This guide builds those ideas through original examples, with checks you can use on SAT® Math modeling questions.
Recognize a constant factor
Consider the sequence 200, 220, 242, 266.2. The differences are 20, 22, and 24.2, so the amount added is not constant. However, each value is 1.1 times the preceding value. This is exponential growth over the listed equal intervals. By contrast, 200, 220, 240, 260 has a constant difference of 20 and describes a linear pattern.
For an exponential model y = a × b^t, a is the value when t = 0 and b is the factor applied whenever t increases by one. In the first sequence, a = 200 and b = 1.1 if the first entry corresponds to t = 0. Then y = 200 × 1.1^t. Substituting t = 2 gives 242, which checks against the table.
The phrase ‘equal intervals’ matters. Ratios from measurements taken at unequal time gaps are not directly comparable as one-period growth factors. A doubling over two hours and a doubling over four hours describe different rates. Label the input values before deciding what a ratio means.
Translate percent change into a base
An increase of r as a decimal multiplies a quantity by 1 + r. A 6% increase therefore uses 1 + 0.06 = 1.06. If an account starts at 500 units and increases by 6% per year under a simplified model, its value after t years is 500 × 1.06^t. After one year, the value is 530, which is 30 more than 500.
A decrease of r multiplies a quantity by 1 − r. A machine worth 8,000 dollars that loses 15% of its value each year follows V = 8,000 × 0.85^t in that simplified depreciation model. The base is 0.85 because 85% remains after each year. Using 0.15 would incorrectly say that only 15% remains.
Conversely, a factor of 1.24 represents 24% growth, while a factor of 0.72 represents 28% decay. Subtract one from a growth factor or subtract a decay factor from one. Then multiply the resulting decimal by 100 to express it as a percentage. Do not read 0.72 as 72% decay.
The decay percentage and the remaining percentage add to 100%. The exponential base uses the remaining fraction.
Interpret the initial value
Substitute zero into the input to identify the initial value. Because b^0 = 1 for a positive base b, y = a × b^0 = a. In y = 350 × 0.9^t, the initial value is 350. The factor 0.9 describes a 10% decrease per input unit, and neither the factor nor the decrease rate is the initial value.
Be careful when the starting time is shifted. The model y = 350 × 0.9^(t − 4) has value 350 when t = 4, not when t = 0. At t = 0 its value is 350 × 0.9^(−4). The coefficient corresponds to the input that makes the exponent zero. This substitution method is more reliable than labeling every leading coefficient ‘the value at time zero.’
Likewise, a model may define t as years since 2020. Then t = 0 means the year 2020, and t = 6 means 2026. Substituting 2026 directly into the exponent would mistake a calendar year for elapsed time. Copy the variable definition beside the equation before using it.
Keep the exponent and time unit aligned
Suppose a population doubles every three hours and begins at 80. If t measures hours, a model is P = 80 × 2^(t/3). At t = 3, one doubling has occurred and the value is 160. At t = 6, two doublings have occurred and the value is 320. The exponent counts the number of three-hour periods.
If a substance halves every five days, a corresponding model is M = M₀ × (1/2)^(d/5), where d is elapsed days. At d = 10, two half-lives have elapsed, so one quarter of the initial amount remains. A half-life refers to repeatedly halving what remains, not repeatedly subtracting half the original amount.
You can rewrite a yearly growth model using months. If A = 1,000 × 1.12^y and m = 12y, then A = 1,000 × 1.12^(m/12). An equivalent monthly base is 1.12^(1/12). It is not exactly 1.01: dividing an annual percentage by twelve ignores compounding. If a problem instead explicitly defines simple interest or a different process, follow that model rather than imposing exponential growth.
Test a model at one complete growth interval. If a quantity is supposed to triple every four days, substituting four days should multiply the initial value by three.
Compare linear and exponential tables
Imagine values of 90, 105, 120, and 135 at inputs 0, 1, 2, and 3. Each step adds 15, so a linear model y = 90 + 15t fits those entries. Now consider 90, 108, 129.6, and 155.52 at the same inputs. Each step multiplies by 1.2, so y = 90 × 1.2^t fits instead.
A common distraction is that both patterns increase. Increasing does not mean exponential. The difference is whether the absolute change or the relative change remains constant. An exponential growth model adds progressively larger amounts when its base is greater than one and its initial value is positive. A positive exponential decay model loses progressively smaller absolute amounts over equal intervals.
If a table contains rounded measurements, differences or ratios may not match perfectly. Read whether the task asks for an exact model, an approximate fit, or the best interpretation. Do not declare a noisy real-world table exactly exponential merely because its entries generally rise. For textbook examples with exact values, repeated ratios provide a clean check.
Growth followed by decay
Increasing a quantity by 20% and then decreasing the result by 20% does not return to the starting value. Starting from 100, the increase gives 120, and the decrease gives 96. Algebraically, the combined factor is 1.2 × 0.8 = 0.96. The second percentage applies to a different starting amount.
If the same two-step cycle repeats, the value after n complete cycles is 100 × 0.96^n. The net change per cycle is a 4% decrease. This is another example of identifying the correct interval: one cycle contains two changes, and the combined factor belongs to that cycle.
To reverse a 20% decrease exactly, divide by 0.8, which is equivalent to multiplying by 1.25. A 25% increase is needed to restore the original value. These percent-change relationships are useful beyond exponential equations because they prevent treating relative changes as interchangeable additive amounts.
Solve for a value or a time
When the question asks for an output at a given time, substitute the time directly. For P = 80 × 2^(t/3), after nine hours P = 80 × 2^3 = 640. Check whether the question asks for the final population or the increase. The increase is 640 − 80 = 560, a different quantity.
When the question asks when a particular value is reached, isolate the exponential expression first. If 80 × 2^(t/3) = 1,280, divide by 80 to get 2^(t/3) = 16. Since 16 = 2^4, t/3 = 4 and t = 12 hours. Recognizing an exact power avoids unnecessary approximation.
For values that are not convenient powers, a graphing calculator can help locate an intersection. Graph the model and a horizontal line at the target value, then interpret the horizontal coordinate in the original time units. If the question requires a whole number of completed periods or asks when a threshold is first exceeded, check the neighboring integers rather than blindly rounding an intersection.
Read a model in context
A model is a stated mathematical description, not a promise that a real process continues forever. A population model with constant percentage growth may fit a limited interval while ignoring resource constraints. A depreciation model may describe expected value without predicting the precise sale price of a particular machine. Answer within the question's assumptions.
The domain can also matter. Time since an experiment began is usually nonnegative in that context, even though the formula might produce a numerical value for negative inputs. If t counts whole annual updates, an integer input may be appropriate; if the model is explicitly continuous over time, fractional inputs may have meaning. The wording determines which interpretation is intended.
Keep units on the output as well as the input. A model giving thousands of bacteria produces a number in thousands. An output of 12 then means 12,000 bacteria, not 12. Many modeling mistakes arise after the algebra is finished because the numerical answer is detached from its defined unit.
Original practice with explained answers
Problem 1: a quantity starts at 240 and decreases by 8% each week. Write a model after w weeks. Answer: Q = 240 × 0.92^w. The retained fraction is 92%, and w counts weekly applications. After two weeks the value is 240 × 0.8464 = 203.136; subtracting 8% of the original value twice would not follow the stated model.
Problem 2: a culture triples every four hours and starts with 50 units. How many units does the model predict after eight hours? Answer: 50 × 3^(8/4) = 450. There are two tripling periods, not eight. The equivalent model with an hourly base would use 3^(1/4), not 3/4.
Problem 3: in N = 600 × 1.05^(t/2), t is measured in days. Interpret 1.05. Answer: the quantity increases by 5% every two days. It does not increase by 5% every day. Setting t = 2 provides a quick check because the exponent becomes one.
Problem 4: a value changes from 320 to 400 over one growth period. What is the percentage increase? The factor is 400/320 = 1.25, so the increase is 25%. Dividing the increase, 80, by the final value, 400, would give 20%, which uses the wrong reference amount.
A short error-checking routine
Before calculating, identify the initial amount, the factor, the length of one factor interval, and the variable's unit. Then test the expression at time zero and at one complete interval. These two substitutions can catch a missing initial value, a reversed growth factor, or an incorrect exponent.
After calculating, ask whether the result moves in the expected direction. A positive quantity undergoing decay should not become larger at a later nonnegative time under the model. If a 5% increase doubles a quantity after one period, a decimal or exponent error is likely. Finally, read the requested quantity again: value, change, factor, rate, and time are different outputs.
Common questions and sources
Is every curved graph exponential? No. Quadratic, rational, and other functions can also be curved. Use the equation, table structure, or stated percentage relationship to identify the model. Does an exponential base of one describe growth? No; with a fixed coefficient it produces a constant value. For the positive models in this guide, a base between zero and one gives decay and a base above one gives growth.
The examples here are original and focus on interpreting nonlinear models. College Board's Math overview describes the current content domains. Pair this topic with our general Math practice guide, and use official Bluebook practice to learn the actual exam tools and presentation.