How to calculate doubling time — step by step
Under exponential growth at a constant rate r, a quantity always takes the same amount of time to double. You find this doubling time exactly with ln(2) ÷ ln(1 + r), or in your head with the rule of 70. Worked example: at 7% growth ≈ 10.24 periods. This topic sits in percentages, suitable for Grade 7 / Year 8.
Quick answer
Doubling time is how long a quantity takes to double at a constant growth rate. Exactly: ln(2) ÷ ln(1 + r). At 7% growth that gives ≈ 10.24 periods. Estimate it fast with the rule of 70: 70 ÷ 7 = 10.
At a glance
| Example | ln(2) ÷ ln(1.07) |
|---|---|
| Method | Exact: ln(2) ÷ ln(1 + r) |
| Steps | ln(2) ÷ ln(1.07) |
| Result | ≈ 10.24 periods |
| Check | Rule of 70: 70 ÷ 7 = 10 |
| Grade level | Grade 7 (ages 12–13) |
Worked example: ln(2) ÷ ln(1.07)
A quantity grows by 7% each period. We work out after how many periods it has doubled.
How to calculate doubling time — the steps
These steps work for any constant growth rate r given in percent.
Step 1 · Start
ln(2) ÷ ln(1 + 7%)Exact formula with r = 7% = 0.07.Step 2 · ln(2)
0.6931 ÷ ln(1.07)ln(2) is a constant ≈ 0.6931.Step 3 · ln(1.07)
0.6931 ÷ 0.0677Evaluate the denominator: ln(1.07) ≈ 0.0677.Step 4 · Result
≈ 10.24After about 10.24 periods the quantity has doubled.Step 5 · Check
70 ÷ 7 ≈ 10Rule of 70 as a sanity check — close to the exact value.
Why the formula works
Doubling means (1 + r) to the power t equals 2. Taking logarithms of both sides turns this into t · ln(1 + r) = ln(2), so t = ln(2) ÷ ln(1 + r). For small r, ln(1 + r) is approximately r, and ln(2) ≈ 0.693 ≈ 0.70 — which gives the rule of thumb doubling time ≈ 70 ÷ percent rate. People often use 72 instead because it divides more easily.