What Is Exponential Growth and Decay?
Exponential growth occurs when a quantity grows by a fixed percentage in each fixed time interval — the larger it gets, the faster it grows. Exponential decay is the mirror image: a quantity shrinks by a fixed percentage per interval, getting smaller at a slower and slower rate. Both are modeled by the same kind of equation, distinguished only by the sign of the growth rate.
The general exponential model is A(t) = A₀ · b^t, where A₀ is the initial amount, b is the growth factor (base), and t is time. If b > 1, the function grows; if 0 < b < 1, it decays.
For example, A(t) = 100 · 2^t doubles every unit of time, while A(t) = 100 · (1/2)^t halves every unit of time. Exponential models describe an enormous range of real phenomena: bacterial populations, radioactive decay, compound interest, depreciation, drug concentration, viral spread, cooling of an object.
The common thread is that the rate of change is proportional to the current amount. This self-reinforcing behavior produces dramatic results — small starting values can grow astoundingly quickly, and large values can shrink surprisingly fast.
The Exponential Growth Formula
The most general exponential growth formula is A(t) = A₀(1 + r)^t, where A₀ is the starting amount, r is the growth rate per period (as a decimal), and t is the number of periods elapsed.
A town has 10,000 people and grows 3% per year. How large after 10 years?
Formula: A(10) = 10,000 · (1 + 0.03)^10 = 10,000 · 1.03^10 ≈ 10,000 · 1.3439 ≈ 13,439.
The town grows by about 34% over the decade — more than just 30% (3% × 10) because each year compounds on the previous total.
≈ 13,439A $5,000 investment earns 6% annual interest. What's it worth after 15 years?
A(15) = 5,000 · 1.06^15 ≈ 5,000 · 2.397 ≈ 11,983.
The investment more than doubles because of compounding.
≈ $11,983Doubling time
Doubling time: The time needed to double depends only on the rate. Solve 2A₀ = A₀(1 + r)^t → t = ln(2)/ln(1 + r). For 3% growth, doubling time ≈ 23.45 years.
The Rule of 72 is a handy approximation: doubling time ≈ 72/r (in percent). For 3% this gives ≈ 24 years — close to the exact answer.
The Exponential Decay Formula
Exponential decay uses the same structure with a negative rate: A(t) = A₀(1 − r)^t, where r is the decay rate per period.
A drug leaves the body at 25% per hour. Starting with 200 mg, how much remains after 6 hours?
A(6) = 200 · (1 − 0.25)^6 = 200 · 0.75^6 = 200 · 0.178 ≈ 35.6 mg.
Less than one-fifth remains after six hours, which is why most medications are dosed multiple times a day.
≈ 35.6 mgA car worth $28,000 loses 15% of its value per year. Worth after 5 years?
A(5) = 28,000 · 0.85^5 ≈ 28,000 · 0.4437 ≈ 12,424.
Cars are a classic example of exponential decay: the dollar loss is largest at the start because the base value is largest.
≈ $12,424Half-life
Half-life: The time for the amount to halve, analogous to doubling time. t_{1/2} = ln(2)/(−ln(1 − r)). For r = 0.25, half-life ≈ 2.41 hours.
Half-life is especially common for radioactive decay, where each isotope has a fixed half-life regardless of starting amount.
Continuous Compounding and the Number e
As the compounding frequency increases, the growth equation approaches a smooth, continuous form based on Euler's number e ≈ 2.71828.
The continuous exponential model is A(t) = A₀ · e^(kt), where k is the continuous growth rate. If k > 0, growth; if k < 0, decay.
This form is essential whenever the rate of change is instantaneously proportional to the current amount — which turns out to be most physical, biological, and financial processes.
Compound interest
Compound interest: Continuously compounded interest at rate r for t years grows as A(t) = P · e^(rt). For daily, monthly, or annual compounding, use A(t) = P · (1 + r/n)^(nt) where n is the number of compounding periods per year.
$1000 at 5% continuously compounded for 10 years:
A(10) = 1000 · e^(0.05 · 10) = 1000 · e^(0.5) ≈ 1000 · 1.6487 ≈ 1648.72.
≈ $1648.72Converting between discrete and continuous rates
Converting between discrete and continuous rates: If (1 + r)^t = e^(kt), then k = ln(1 + r). For r = 0.03, k = ln(1.03) ≈ 0.02956.
The continuous rate is slightly smaller than the discrete rate, but very close for small rates.
Half-Life and Doubling Time
Half-life is the time required for a decaying quantity to reduce by half. For continuous decay A(t) = A₀ · e^(−kt), solving A₀/2 = A₀ · e^(−k · t_{1/2}) gives t_{1/2} = ln(2)/k.
Doubling time for continuous growth A(t) = A₀ · e^(kt): t_2 = ln(2)/k.
Notice they share the same formula with k being the (positive) growth or decay rate.
Carbon-14 has a half-life of 5,730 years.
Decay rate: k = ln(2)/5730 ≈ 0.000121 per year.
If a sample starts with 100 grams, how much remains after 10,000 years?
A(10,000) = 100 · e^(−0.000121 · 10,000) = 100 · e^(−1.21) ≈ 100 · 0.2981 ≈ 29.81 grams.
≈ 29.81 gramsBacteria grow continuously at 30% per hour.
Doubling time: t_2 = ln(2)/0.3 ≈ 2.31 hours.
Starting with 500 bacteria, the population reaches 1000 after 2.31 hours, 2000 after 4.62 hours, 4000 after 6.93 hours, and so on.
t_2 ≈ 2.31 hoursThe doubling-and-half-life concepts are powerful because they don't depend on the starting amount — radiocarbon dating uses exactly this property to estimate the age of organic materials.
Solving Exponential Equations
Many real-world exponential problems reduce to solving for t given the amount. The key tool is the logarithm, which is the inverse of exponentiation.
- General technique
Isolate the exponential expression, then take
ln(orlog) of both sides.
A $5,000 investment grows at 6% compounded continuously. How long to reach $10,000?
10,000 = 5,000 · e^(0.06t).
Divide: 2 = e^(0.06t).
Take ln: ln(2) = 0.06t.
Solve: t = ln(2)/0.06 ≈ 11.55 years.
t ≈ 11.55 yearsA population grows from 1,000 to 1,500 in 4 years (continuous growth). Find k.
1,500 = 1,000 · e^(4k).
Divide: 1.5 = e^(4k).
Take ln: ln(1.5) = 4k.
Solve: k = ln(1.5)/4 ≈ 0.1014 or 10.14% per year.
k ≈ 0.1014A fossil has 12% of its original carbon-14. How old?
Using half-life 5730 years, k = ln(2)/5730.
Solve 0.12 = e^(−kt): ln(0.12) = −kt, so t = −ln(0.12)/k ≈ 17,532 years.
t ≈ 17,532 yearsIf you're working through exponential problems and want to double-check your work, scan them with Solver AI — it will show every algebraic and logarithmic step with explanations.