Exponential Growth & Decay: When Change Compounds
Model exponential growth and decay — y = a(1 ± r)^t, reading the base, compound interest, and half-life — with a clear graph and worked SAT and ACT problems.
The Short Version
- Exponential change multiplies by a fixed factor each step (constant percent), unlike linear's constant amount.
- Model it with y = a(1 + r)^t for growth and y = a(1 − r)^t for decay.
- Read the base: greater than 1 means growth, between 0 and 1 means decay.
- Powers compound interest and half-life problems — an SAT/ACT and Algebra II topic, beyond the SSAT.
Linear growth adds: save $50 a month and your balance climbs by the same $50 each time. Exponential growth multiplies: earn 5% interest and each month's gain is bigger than the last, because you earn interest on your interest. That shift from adding a fixed amount to multiplying by a fixed percentage is the whole idea behind exponential models — and it governs money, populations, and radioactive decay alike.
This guide contrasts linear and exponential change, builds the growth/decay formula, and applies it to interest and half-life, with worked and practice problems matched to real test difficulty at Northside Tutoring.
Why Exponential Models Matter
Exponential growth and decay are favorites on the SAT and ACT because they connect algebra to the real world: savings accounts, population models, depreciation, and decay. The questions test whether you can read a percentage rate out of a formula and interpret what the numbers mean. It's Algebra II material beyond the SSAT.
Linear vs. Exponential
The defining question: does the quantity change by a constant amount or a constant percent? Constant amount is linear (y = mx + b). Constant percent is exponential. "Grows by 5 each year" is linear; "grows by 5% each year" is exponential.
The Growth & Decay Formula
The standard model uses a starting amount a, a rate r (as a decimal), and time t:
For 8% growth, the base is 1.08; for 8% decay, it's 0.92. The starting amount a is the value when t = 0.
Reading the Base
You can tell growth from decay at a glance by the base b in y = a·b^t:
| Base b | Behavior | Percent rate |
|---|---|---|
| b > 1 (e.g., 1.08) | growth | b − 1 = 8% increase |
| 0 < b < 1 (e.g., 0.92) | decay | 1 − b = 8% decrease |
What the Graphs Look Like
Growth curves upward ever more steeply; decay falls fast then levels off, approaching zero.
Interest & Half-Life
Two classic applications:
- Compound interest: A = P(1 + r/n)^(nt), where n is the number of times interest compounds per year.
- Half-life: a decay where the amount halves each period: y = a(½)^(t/h), with h the half-life.
Convert the percent first
The rate r in the formula is a decimal: 6% becomes 0.06. Plugging in 6 instead of 0.06 is a common and costly error.
Where You'll See This — Test by Test
No reference sheet provides these models. The SAT and ACT test reading the base for growth vs. decay, finding the percent rate, and evaluating the formula. It's Algebra II content beyond the SSAT.
Digital SAT
Tests interpreting exponential models in context: identifying growth vs. decay, the rate, and the starting value.
Explore SAT Tutoring → College AdmissionsACT
Tests evaluating exponential expressions, compound interest, and growth/decay word problems.
Explore ACT Tutoring → Independent School AdmissionsSSAT
Not on the SSAT — Algebra II material. Build exponent fluency with earlier prep first.
Explore SSAT Tutoring → K-12 CurriculumAlgebra II
A core Algebra II topic and the foundation for logarithms and continuous growth.
Explore Algebra Tutoring →Watch the Lesson
Sometimes a diagram needs a voice. In the short video below, one of our Northside tutors walks through the core idea and works through test-style problems in real time.
Growth & Decay — In Plain English
A live walkthrough from our tutoring team.
— Featuring a Northside Tutoring instructor
Worked Example Problems
These problems are calibrated to the difficulty you'll actually see on test day. Try each one before opening the solution.
Does y = 300(1.07)^t represent growth or decay, and at what rate?
Show solution
The base 1.07 is greater than 1, so it's growth.
Rate = 1.07 − 1 = 0.07 = 7% per period.
A $500 investment grows 4% per year. What is its value after 3 years?
Show solution
y = 500(1.04)³ = 500(1.124864) ≈ 562.43.
A car worth $20,000 loses 15% of its value each year. Write a model for its value after t years.
Show solution
Decay at 15%: base = 1 − 0.15 = 0.85.
y = 20000(0.85)^t.
A sample of 80 mg has a half-life of 5 years. How much remains after 10 years?
Show solution
10 years = 2 half-lives. Halve twice: 80 → 40 → 20.
In y = 1200(0.9)^t, what does the 1200 represent?
Show solution
The value when t = 0 — the starting amount. (The base 0.9 means 10% decay per period.)
Common Mistakes to Avoid
Three traps that catch students every year
- Confusing constant amount with constant percent. "Grows by 5" is linear; "grows by 5%" is exponential.
- Using the percent instead of the decimal. The rate r must be a decimal — 6% is 0.06, not 6.
- Misreading the base. Base > 1 is growth; a base between 0 and 1 is decay. The percent is the distance from 1.
Practice Problems — You Try
Three problems below. Work each before checking the solution.
Is y = 50(0.8)^t growth or decay, and at what rate?
Show solution
Base 0.8 is between 0 and 1 → decay. Rate = 1 − 0.8 = 20%.
A population of 1,000 grows 10% per year. What is it after 2 years?
Show solution
1000(1.10)² = 1000(1.21) = 1210.
A substance decays to half every 3 hours. Starting at 240 g, how much remains after 9 hours?
Show solution
9 hours = 3 half-lives. Halve three times: 240 → 120 → 60 → 30.
The Northside Method — How We Teach This 1-on-1
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