PSAT/NMSQT Math: Exponential Growth and Decay
Exponential models describe quantities that change by a constant percentage over equal time intervals, rather than by a constant amount. This free, complete lesson covers writing growth and decay models from a percentage change, doubling time and half-life, compound interest, and comparing the speed of different growth rates. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the linked quiz below.
1. Writing Growth and Decay Models from Percentage Change
An exponential model has the form y = a(1 + r)t for growth, or y = a(1 − r)t for decay, where a is the initial value, r is the rate as a decimal, and t is time. Unlike a linear model, which adds the same fixed amount each period, an exponential model multiplies by the same factor each period.
Worked Example: An investment of $2,000 loses 4% of its value every year. Write a model for its value V after t years.
V = 2000(1 − 0.04)t = 2000(0.96)t
1. Which of the following represents exponential growth, rather than linear growth or decay?
Explanation: A base greater than 1 raised to the power t (here, 1.08) represents exponential growth; a base between 0 and 1 represents decay, and the other two options are linear.
2. A town's population of 12,000 grows by 3% each year. Write a model for the population P after t years.
Explanation: Growth of 3% per year means multiplying by (1 + 0.03) = 1.03 each year, giving P = 12000(1.03)t.
3. Which model correctly represents a car worth $28,000 depreciating (losing value) by 12% per year?
Explanation: Losing 12% each year means multiplying by (1 − 0.12) = 0.88 each year.
2. Doubling Time and Half-Life
Doubling time is the fixed period it takes a growing quantity to double, and half-life is the fixed period it takes a decaying quantity to reduce by half. Both let you skip the full exponential formula: after n full doubling periods, a growing quantity is multiplied by 2n; after n full half-lives, a decaying quantity is multiplied by (1/2)n.
Worked Example: A bacteria culture doubles every 4 hours. If it starts with 300 bacteria, how many are present after 12 hours?
12 hours is exactly 3 doubling periods (12/4 = 3):
300 × 2³ = 300 × 8 = 2,400 bacteria
1. A population doubles every 6 years. If it starts at 500, what is the population after 18 years?
Explanation: 18 years is 3 doubling periods (18/6 = 3): 500 × 2³ = 500 × 8 = 4,000.
2. A radioactive sample starts at 80 grams and has a half-life of 5 years. How much remains after 20 years?
Explanation: 20 years is 4 half-lives (20/5 = 4): 80 × (1/2)⁴ = 80 × 1/16 = 5 grams.
3. A quantity's growth is modeled by Q = 40 · 2t/3, where t is in days. What does this equation tell you about the doubling time?
Explanation: The exponent t/3 reaches 1 (one full doubling) every time t increases by 3, so the doubling time is 3 days.
3. Compound Interest and Growth Rate
Compound interest is a direct application of exponential growth: A = P(1 + r)t, where P is the principal (starting amount), r is the annual interest rate as a decimal, and A is the balance after t years. Given any three of these values, you can solve algebraically for the fourth.
Worked Example: $5,000 is invested at an annual rate of 4%, compounded yearly. Find the balance after 3 years.
A = 5000(1.04)³ ≈ $5,624.32
1. $3,000 is invested at 5% annual interest, compounded yearly. What is the balance after 2 years?
Explanation: A = 3000(1.05)² = 3000(1.1025) = $3,307.50.
2. An account modeled by A = 1000(1 + r)t grows to $1,210 after 2 years. Set up the equation to solve for the annual rate r.
Explanation: Substituting A = 1210, P = 1000, and t = 2 directly into A = P(1 + r)t gives 1210 = 1000(1 + r)².
3. Using 1210 = 1000(1 + r)² from the previous problem, what is the annual interest rate r?
Explanation: 1.21 = (1 + r)² → √1.21 = 1 + r → 1.1 = 1 + r → r = 0.10 = 10%.
4. Comparing Growth Rates
When comparing two exponential functions, the one with the larger base (for growth, base > 1) grows faster, and eventually overtakes any function with a smaller base, no matter how large the starting value of the slower-growing function is. For an exponential function y = a · bx, increasing x by exactly 1 always multiplies y by the base b.
Worked Example: Between f(x) = 100(1.05)x and g(x) = 10(1.20)x, which eventually grows faster, regardless of starting value?
g(x) has the larger base (1.20 > 1.05), so g(x) eventually grows faster and overtakes f(x)
1. Which function grows fastest as x increases without bound?
Explanation: The function with the largest base, 3, will eventually outgrow every other option, regardless of their starting values.
2. For the function y = 20(1.15)x, what happens to y each time x increases by 1?
Explanation: Increasing the exponent by 1 always multiplies the entire function value by the base, here 1.15.
3. A substance decays continuously according to Q = Q₀e−0.05t. What fraction of the original amount remains when the exponent equals −1 (i.e., 0.05t = 1)?
Explanation: When the exponent equals −1, Q = Q₀e−1 = Q₀/e, and since e ≈ 2.718, this fraction is about 0.37, or 37%.
Common Mistakes to Avoid
- Using the percentage itself as the base instead of (1 ± r). A 5% growth rate gives a base of 1.05, not 5 or 0.05.
- Confusing growth and decay bases. A base greater than 1 means growth; a base between 0 and 1 (but greater than 0) means decay.
- Miscounting the number of doubling periods or half-lives. Always divide the total elapsed time by the length of one period to get a whole (or fractional) number of periods before applying the multiplier.
- Assuming a larger starting value always stays larger. A function with a smaller starting value but a larger base will eventually overtake one with a bigger start but a smaller base.
- Forgetting to take a root when solving for a rate inside an exponent, such as needing a square root to undo squaring the growth factor over 2 years.
Frequently Asked Questions
How do I tell exponential growth apart from linear growth on the PSAT?
Exponential growth multiplies by a constant factor each period (fast, accelerating change), while linear growth adds a constant amount each period (steady, constant change). Check whether the function has the input variable in the exponent, that's the exponential signal.
Do I need to memorize a separate formula for half-life problems?
No, half-life is just decay with a base of 1/2, applied once for every full half-life period that has passed. You can always write it as (1/2)t/h, where h is the half-life length, rather than memorizing a separate formula.
Where can I practice more problems like these?
The Exponential Growth and Decay quiz in the PSAT/NMSQT Math Question Bank includes additional original problems on this topic, along with quizzes covering every other Heart of Algebra, Advanced Math, Problem Solving and Data Analysis, and Geometry and Trigonometry skill tested on the PSAT/NMSQT. You can also browse the full PSAT/NMSQT Math Question Bank.