Last updated: 2026-07-01

Exponential growth and decay

A quantity grows or decays exponentially when its rate of change is proportional to its current size. The bigger it is, the faster it changes. This pattern describes populations, compound balances, radioactive material, drug concentrations, and the early spread of an epidemic. The single model behind all of them is:

N = N₀ (1 ± r)^t

Here N₀ is the initial amount, r is the per-period rate, t is the elapsed time, and N is the amount after that time. A plus sign gives growth; a minus sign gives decay.

Growth versus decay

The only difference between growth and decay is the sign of the rate. With growth the multiplier (1 + r) is greater than 1, so each period the quantity is scaled up and the curve rises ever more steeply. With decay the multiplier (1 − r) lies between 0 and 1, so each period the quantity is scaled down and the curve falls toward — but never quite reaches — zero.

Discrete versus continuous

There are two common ways to apply the rate.

The discrete model, N₀ (1 ± r)^t, applies the rate once per whole period. It fits situations that change in steps: interest credited once a year, a population censused each generation, a balance updated each month.

The continuous model, N₀ · e^(kt), applies the rate at every instant, where k is the continuous rate and e ≈ 2.71828. It fits processes that change smoothly all the time: radioactive decay, continuously compounded interest, first-order chemical reactions. For the same nominal rate the continuous model grows a little faster, because each tiny gain immediately begins compounding.

Doubling time and half-life

Because the rate is a fixed percentage, a growing quantity takes the same time to double no matter how large it already is, and a decaying quantity takes the same time to halve. These constants are the doubling time and the half-life.

For the discrete model:

• Doubling time = ln 2 ÷ ln(1 + r)
• Half-life = −ln 2 ÷ ln(1 − r)

For the continuous model both reduce to ln 2 ÷ k, since the magnitude of the rate is the same in either direction.

A handy shortcut is the rule of 70: divide 70 by the percentage rate to estimate the doubling time. At 5% per year, doubling takes roughly 70 ÷ 5 = 14 years — close to the exact 14.21 years.

Worked example

Problem: A town of 1,000 people grows at 5% per year. How many people are there after 10 years, and how long until the town doubles?

Using the discrete growth model:

N = 1000 × (1 + 0.05)^10 = 1000 × 1.62889 ≈ 1,629 people

Doubling time = ln 2 ÷ ln(1.05) ≈ 0.6931 ÷ 0.04879 ≈ 14.21 years

So after a decade the population is about 1,629, and at this rate it will reach 2,000 in just over fourteen years.

A decay example

Problem: A 200 mg dose of a drug decays continuously with a rate of 0.231 per hour. How much remains after 6 hours, and what is its half-life?

N = 200 × e^(−0.231 × 6) = 200 × e^(−1.386) ≈ 50 mg

Half-life = ln 2 ÷ 0.231 ≈ 3 hours

The 6-hour result of 50 mg is exactly a quarter of the dose, which makes sense: 6 hours is two half-lives, so the amount has halved twice.

Relationship to compound interest

Exponential growth is the mathematical engine behind compound interest. The discrete model with r as the periodic interest rate is identical to the compound-interest formula, and the continuous model corresponds to continuously compounded interest. This calculator keeps the framing general so it applies equally to biology, physics, and finance.