Exponential growth and decay describe any quantity that changes at a rate proportional to its current size — from savings accounts to bacterial colonies, radioactive isotopes to viral spread. This calculator implements the exact mathematical formula for both continuous and discrete models, lets you solve for any of the four variables, and shows the complete algebraic working so you can follow every step.
The core formula
Continuous exponential growth is governed by the ordinary differential equation:
dN/dt = r · N(t)
Integrating from 0 to t gives the closed-form solution:
N(t) = N₀ · e^(r · t)
where N₀ is the initial quantity, r is the continuous growth rate (per unit time), t is elapsed time, and e ≈ 2.71828 is Euler’s number. When r is positive the quantity grows; when r is negative it decays.
Discrete (periodic) compounding — the form used in finance — credits growth at the end of each compounding period:
N(t) = N₀ · (1 + r/n)^(n · t)
where r is the nominal annual rate and n is the number of compounding periods per year (1 = annual, 4 = quarterly, 12 = monthly, 365 = daily). As n → ∞ the discrete formula converges exactly to the continuous formula.
Solve-for-variable
The calculator can find any one of the four quantities — N(t), N₀, r, or t — given the other three. The rearranged formulas used internally are:
| Unknown | Continuous | Discrete |
|---|---|---|
| N(t) | N₀ · e^(rt) | N₀ · (1+r/n)^(nt) |
| N₀ | N(t) / e^(rt) | N(t) / (1+r/n)^(nt) |
| r | ln(N(t)/N₀) / t | n · [(N(t)/N₀)^(1/(nt)) − 1] |
| t | ln(N(t)/N₀) / r | ln(N(t)/N₀) / (n · ln(1+r/n)) |
Doubling time and half-life
Two derived quantities follow automatically from the rate:
- Doubling time T₂ = ln(2) / r (continuous) — the time for any growing quantity to double in size. The popular “Rule of 70” approximates this as 70/r%, which works because ln(2) ≈ 0.693 and 100 × ln(2) ≈ 69.3 ≈ 70.
- Half-life T½ = ln(2) / |r| (continuous) — the time for a decaying quantity to halve, used in nuclear physics, pharmacokinetics, and carbon dating.
Worked example — savings at 7% continuous growth
Start with $10,000 growing continuously at 7% per year for 20 years:
N(20) = 10,000 · e^(0.07 × 20) = 10,000 · e^1.4 ≈ $40,552
Compare to monthly discrete compounding at 7% nominal (n = 12):
N(20) = 10,000 · (1 + 0.07/12)^(12×20) ≈ $40,096
The continuous model gives a slightly higher result because interest compounds at every infinitesimal moment. The doubling time at 7% continuous is ln(2)/0.07 ≈ 9.90 years (Rule of 70 estimate: 10 years — remarkably close).
Worked example — radioactive decay (half-life to rate)
Carbon-14 has a measured half-life of 5,730 years. What fraction remains after 2,000 years?
First extract r: r = −ln(2) / 5730 ≈ −0.0001210 per year.
Then: N(2000) = N₀ · e^(−0.0001210 × 2000) = N₀ · e^(−0.2421) ≈ 0.7849 · N₀
About 78.5% of the original carbon-14 remains after 2,000 years. Enter r = −0.01210 (as % per century) and t = 20 to reproduce this result with the calculator.
Formula reference note
The continuous formula N(t) = N₀ · e^(rt) is the standard result from Malthusian population dynamics (Thomas Malthus, 1798) and the Verhulst logistic model’s unconstrained limit. It appears in nuclear decay (Bateman equations), pharmacokinetic elimination, heat transfer (Newton’s law of cooling), and finance (Black-Scholes continuous compounding). The discrete form matches the compound interest formula standardised by actuaries in the 17th century and codified in modern accounting via the Annual Equivalent Rate (AER) definition.