"Exponential growth and decay" is searched about 2,800 times a month in the United States, by Algebra 1 and 2 students and by the biology, chemistry and economics students who meet the same formula under other names (compound interest, half-life, population growth). One formula covers all of it. This page gives the formula and the two factors, how to read a, b and r from equations, tables and words, the half-life and doubling-time forms, compound interest, the continuous version with e, eight worked examples with the arithmetic shown, the graphs, the mistakes, and a printable formula sheet with practice.
Download the exponential growth and decay sheet with practice (PDF) (A4, the formulas, the factor table and the worked examples on page 1, the half-life and compound interest cases, the graphs and the practice set with answers on page 2, free).
The formula
A quantity changes exponentially when it is multiplied by the same factor every period (every year, every hour, every generation), as opposed to linear change, where the same amount is added.
y = a · bᵗ
- a: the starting amount (the value when t = 0)
- b: the growth or decay factor, the number the quantity is multiplied by each period
- t: the number of periods
- y: the amount after t periods
The factor comes from the percentage rate r (as a decimal):
| Growth | Decay | |
|---|---|---|
| Formula | y = a(1 + r)ᵗ | y = a(1 − r)ᵗ |
| Factor b | 1 + r, bigger than 1 | 1 − r, between 0 and 1 |
| 5% a period | b = 1.05 | b = 0.95 |
| 20% a period | b = 1.20 | b = 0.80 |
| 100% (doubling) | b = 2 | |
| 50% (halving) | b = 0.5 | |
| Graph | rises, ever steeper | falls, ever flatter, toward 0 |
Adding 8% means keeping 100% and adding 8%: ×1.08. Losing 15% means keeping 85%: ×0.85. That is where the 1 comes from.
Reading a, b and r
From an equation. y = 500(1.03)ᵗ: a = 500, b = 1.03, so r = 0.03, growth of 3% per period. y = 20(0.6)ᵗ: a = 20, b = 0.6, r = 0.4, decay of 40% per period.
From a table with equal steps in t: divide each value by the one before.
| t | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| y | 200 | 240 | 288 | 345.6 |
240 ÷ 200 = 1.2; 288 ÷ 240 = 1.2; constant ratio, so exponential with b = 1.2 (20% growth), a = 200, y = 200(1.2)ᵗ. A table with constant differences (200, 240, 280, 320) is linear instead.
From words. "A town of 12,000 grows 2% a year": a = 12,000, b = 1.02. "A car bought for $20,000 loses 15% of its value each year": a = 20,000, b = 0.85. "Bacteria double every 20 minutes": b = 2 per 20-minute period. "Half-life 5 hours": b = 0.5 per 5-hour period.
Half-life and doubling time
When the period in the question is not the unit of time (half-life 5 hours, but you want the amount after 12 hours), write the exponent as t ÷ T, the number of half-lives or doublings that have passed:
- Decay with half-life T: y = a · (½)^(t/T)
- Growth with doubling time T: y = a · 2^(t/T)
Caffeine, 80 mg, half-life 5 h, after 12 h: y = 80 × (½)^(12/5) = 80 × (½)^2.4 = 80 × 0.1895 = 15.2 mg. After 15 h (three half-lives): 80 × ⅛ = 10 mg.
Bacteria, 500 cells, doubling every 20 min, after 2 h (120 min, six doublings): y = 500 × 2⁶ = 500 × 64 = 32,000.
Carbon-14 has a half-life of 5,730 years; a sample with 25% of its original carbon-14 has been through two half-lives, so it is about 11,460 years old.
Compound interest is exponential growth
| Compounding | Formula | $5,000 at 4% for 10 years |
|---|---|---|
| Yearly | A = P(1 + r)ᵗ | 5,000 × 1.04¹⁰ = 5,000 × 1.4802 = $7,401.22 |
| n times a year | A = P(1 + r/n)ⁿᵗ | monthly: 5,000 × (1 + 0.04/12)¹²⁰ = $7,454.16 |
| Continuously | A = Peʳᵗ | 5,000 × e^0.4 = 5,000 × 1.4918 = $7,459.12 |
More frequent compounding gives slightly more, and continuous compounding is the limit. Simple interest, by contrast, is linear: 5,000 + 5,000 × 0.04 × 10 = $7,000.
The continuous form: y = aeᵏᵗ
In Algebra 2, A-level and the sciences the formula is often written with the number e ≈ 2.718: y = a·eᵏᵗ, with k the continuous rate (positive for growth, negative for decay). It is the same curve: eᵏ = b, so y = 300e^(0.05t) has b = e^0.05 = 1.0513, about 5.13% growth per period. Radioactive decay is usually written N = N₀e^(−λt), with the decay constant λ = ln 2 ÷ T (T the half-life). Solving for t needs logarithms: t = ln(y ÷ a) ÷ k.
Eight worked examples
- Population. A town of 12,000 grows 2% a year. After 15 years: 12,000 × 1.02¹⁵ = 12,000 × 1.3459 = 16,150.
- Car value. $20,000, losing 15% a year. After 5 years: 20,000 × 0.85⁵ = 20,000 × 0.4437 = $8,874.
- Savings. $5,000 at 4% compounded yearly for 10 years: $7,401.22 (above).
- Caffeine. 80 mg, half-life 5 h, after 15 h: 10 mg.
- Bacteria. 500, doubling every 20 min, after 2 h: 32,000.
- Reading a table. 1,000; 900; 810; 729: each ÷ the previous = 0.9, so 10% decay, y = 1,000(0.9)ᵗ. After 10 periods: 1,000 × 0.9¹⁰ = 349.
- Finding the rate. A plant grows from 10 cm to 14.4 cm in 2 weeks exponentially. b² = 1.44, b = 1.2: 20% a week.
- Finding the time (Algebra 2). How long for $1,000 at 6% to double? 2 = 1.06ᵗ, t = ln 2 ÷ ln 1.06 = 0.693 ÷ 0.0583 = 11.9 years. (The "rule of 72": 72 ÷ 6 = 12, a quick estimate.)
The graphs
Both graphs pass through (0, a), the starting amount. Growth (b > 1) rises to the right, slowly at first and then steeply, and falls toward the x-axis to the left. Decay (0 < b < 1) falls to the right, steeply at first and then ever more gently, never reaching zero: the x-axis is a horizontal asymptote. Neither graph ever touches the x-axis, since a·bᵗ is never zero for positive a. The y-intercept is a; the point (1, ab) gives the factor.
The mistakes that cost marks
- Using r instead of 1 ± r as the base: y = 500(0.03)ᵗ instead of 500(1.03)ᵗ.
- Writing the percentage as a whole number: 1 + 8 instead of 1 + 0.08.
- Adding instead of multiplying across periods (20% growth for 3 years is × 1.2³ = 1.728, not + 60%).
- Using t in hours when the factor is per 5 hours. Divide: t ÷ 5.
- Rounding 1.04¹⁰ to 1.4 and losing the cents. Keep four decimal places until the end.
- Confusing a table with constant differences (linear) with one with constant ratios (exponential).
- Saying a decaying quantity "reaches zero". It approaches zero.
Where it sits in each curriculum
| Curriculum | When | What is expected |
|---|---|---|
| US Common Core | Algebra 1 (HSF-LE.A.1, A.2: distinguish linear from exponential, construct exponential functions from tables and descriptions; HSF-LE.B.5: interpret a, b in context; HSF-IF.C.8b), Algebra 2 (HSF-LE.A.4: solve for t with logarithms; the e form) | formula, factor, tables, half-life; logs in Algebra 2 |
| England | GCSE Higher (compound interest and depreciation, exponential growth and decay, graphs of y = kˣ), A-level (eˣ, ln, modelling with y = aeᵏᵗ) | compound interest and decay at GCSE, e and ln at A-level |
| Australia | Year 10 (compound interest, exponential functions and their graphs), Year 11 and 12 Mathematical Methods (exponential and logarithmic functions, growth and decay models) | modelling in Year 11 and 12 |
| India (CBSE) | Class 8 (Comparing Quantities: compound interest and its applications to growth and depreciation), Class 11 and 12 (exponential and logarithmic functions) | compound interest as the model in Class 8 |
Practice (answers in the PDF)
- Identify a, b and r, and say whether it is growth or decay: (a) y = 2,500(1.07)ᵗ; (b) y = 64(0.5)ᵗ; (c) y = 90(0.92)ᵗ.
- A city of 850,000 grows 1.5% a year. What is its population after 8 years?
- A laptop costs $1,200 and loses 20% of its value each year. What is it worth after 3 years?
- A medicine has a half-life of 6 hours. If 200 mg is taken, how much remains after 18 hours? After 15 hours?
- A table shows 50, 60, 72, 86.4 at t = 0, 1, 2, 3. Is it exponential? Write the equation.
- $3,000 is invested at 5% compounded yearly. Write the formula and find the balance after 6 years.
Download the formula sheet and the practice set (PDF)
Where to get help
A 1-on-1 online maths class with a tutor who holds a master's in mathematics and teaches Algebra 1 and 2, GCSE and A-level maths, the Australian Years 10 to 12 and CBSE Class 8, 11 and 12 every week: exponential change taught by the multiply-each-period idea first, then the formula, with the "keep 100% and add" step for the base rehearsed until 15% decay is 0.85 without thinking. The Algebra 2 page lists the units; the free 30-minute first class is a real lesson on this week's homework. Related pages: the square root function and the algebra formulas list.
Questions parents ask
1What is the formula for exponential growth and decay?
2What is the difference between exponential growth and exponential decay?
3How do you find the growth or decay factor?
4What is half-life, and how do you calculate with it?
5Is compound interest exponential growth?
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