Exponential Growth & Decay Calculator (Free)
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An exponential growth calculator uses x(t) = x0(1 + r)t, where x0 is the starting amount, r the rate per time step and t the number of steps. For example, 1,000 growing 5% a year for 10 years becomes 1,000 × 1.0510 ≈ 1,628.89. A negative rate gives exponential decay.
Show steps
- Write the rate as a decimal: r = 5% ÷ 100 = 0.05. Each year multiplies the amount by 1 + r = 1.05.
- Apply that factor t = 10 times: (1.05)^10 ≈ 1.62889463.
- Multiply by the starting amount: 1,000 × 1.62889463 ≈ 1,628.89.
- Total change: 1.62889463 − 1 = +62.89%.
- Doubling time: solve (growth factor)^t = 2, so t = ln 2 ÷ ln(1 + r) = 0.693147 ÷ 0.04879016 ≈ 14.2067 years.
- The same growth as a continuous rate: k = ln(1 + r) = +4.879% per year, and x₀ × e^(kt) gives the same 1,628.89.
Table of values
| t | x(t) | Change | × start |
|---|---|---|---|
| 0 | 1,000 | 0% | 1 |
| 1 | 1,050 | +5% | 1.05 |
| 2 | 1,102.5 | +10.25% | 1.1025 |
| 3 | 1,157.63 | +15.76% | 1.157625 |
| 4 | 1,215.51 | +21.55% | 1.215506 |
| 5 | 1,276.28 | +27.63% | 1.276282 |
| 6 | 1,340.1 | +34.01% | 1.340096 |
| 7 | 1,407.1 | +40.71% | 1.4071 |
| 8 | 1,477.46 | +47.75% | 1.477455 |
| 9 | 1,551.33 | +55.13% | 1.551328 |
| 10 | 1,628.89 | +62.89% | 1.628895 |
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How to use the exponential growth calculator
- Pick what you want to find: the final value x(t), the rate, the time, the initial value x0, or the result of a doubling time or half-life.
- Choose the model: periodic, x0(1 + r)t, or continuous, x0ekt.
- Enter the other values. Use a negative rate, such as −10, for decay, and pick the time unit your rate uses.
- Read the answer, then open Show steps, the graph and the table of values. The example buttons load a population, bacteria, carbon-14 and car-value problem.
Exponential growth and decay formulas
Periodic model: x(t) = x0(1 + r)t
- x0 is the initial value, at t = 0.
- r is the rate per time step as a decimal: 5% growth is 0.05, 10% decay is −0.10.
- t is the time, in the same unit as the rate.
- 1 + r is the growth factor: above 1 means growth, between 0 and 1 means decay.
Continuous model: x(t) = x0ekt, where k is the continuous rate and e ≈ 2.71828. The two models describe the same curve when k = ln(1 + r). A 5% periodic rate equals a continuous rate of about 4.879%, and a continuous rate of 5% equals a periodic rate of about 5.1271%.
Worked example: periodic vs continuous growth
Start with 1,000 and a rate of 5% for 10 time units. Periodic: 1,000 × 1.0510 ≈ 1,628.89. Continuous: 1,000 × e0.05 × 10 = 1,000 × e0.5 ≈ 1,648.72. The same percentage gives a larger result in the continuous model, because the growth is added at every instant.
Worked example: exponential decay
1,000 decreasing 10% per period for 3 periods: 1,000 × 0.93 = 729. The total drop is 27.1%, not 30%, because each period takes 10% of what is left, not of the original amount.
How to solve for the rate, time or initial value
- Rate: r = (x(t) ÷ x0)1/t − 1, or k = ln(x(t) ÷ x0) ÷ t. Growing from 1,000 to 2,000 in 10 years takes r ≈ 7.1773% a year.
- Time: t = ln(x(t) ÷ x0) ÷ ln(1 + r), or ln(x(t) ÷ x0) ÷ k. At 5% a year, 1,000 reaches 2,000 after about 14.2067 years.
- Initial value: x0 = x(t) ÷ (1 + r)t. Divide out the growth to go back in time.
Doubling time and half-life
The doubling time is how long a growing amount takes to double: t2 = ln 2 ÷ ln(1 + r), or ln 2 ÷ k. The half-life is how long a decaying amount takes to halve: t½ = ln 2 ÷ |ln(1 + r)|. For a quick estimate, divide 70 (or 72) by the percentage rate.
| Rate per period | Doubling time or half-life |
|---|---|
| +1% | 69.66 periods to double |
| +2% | 35.00 periods to double |
| +3% | 23.45 periods to double |
| +5% | 14.21 periods to double |
| +7% | 10.24 periods to double |
| +10% | 7.27 periods to double |
| −5% | 13.51 periods to halve |
| −10% | 6.58 periods to halve |
| −20% | 3.11 periods to halve |
| −50% | 1 period to halve |
When you know the doubling time or half-life instead of the rate, use x(t) = x0 × 2t/T or x(t) = x0 × (1/2)t/T, where T is the doubling time or half-life.
Real-world examples
Population growth
A town of 50,000 grows 2% a year. After 10 years: 50,000 × 1.0210 ≈ 60,949.72, so about 60,950 people. At that rate the population doubles every 35.00 years.
Bacteria doubling
A culture starts with 100 bacteria and doubles every 20 minutes. Three hours is 180 ÷ 20 = 9 doublings, so the count is 100 × 29 = 51,200.
Radioactive decay: carbon-14
Carbon-14 has a half-life of 5,730 years (NOAA). After 10,000 years, a 100-gram sample has passed 10,000 ÷ 5,730 ≈ 1.745 half-lives, leaving 100 × (1/2)1.745 ≈ 29.83 grams. As a continuous rate, k = −ln 2 ÷ 5,730 ≈ −0.000121 per year.
Depreciation
A car bought for $30,000 that loses 15% of its value each year is worth 30,000 × 0.855 ≈ $13,311.16 after 5 years. Money growing at a fixed interest rate follows the same model; see the compound interest calculator.
Exponential vs linear growth
Linear growth adds the same amount each step; exponential growth multiplies by the same factor. Starting at 1,000, adding 50 a year gives 1,500 after 10 years and 2,500 after 30. Growing 5% a year gives 1,628.89 after 10 years and about 4,321.94 after 30. The gap keeps widening because each step’s growth is based on a bigger amount.
Common mistakes
- Entering 5% as 5 instead of 0.05 in the formula (the calculator takes 5 and converts it for you).
- Using a yearly rate with time in months. Match the units, or convert the rate first.
- Treating decay as subtraction: 10% decay for 3 periods leaves 72.9%, not 70%.
- Mixing up r and k: x0(1 + r)t and x0ekt agree only when k = ln(1 + r).
- Assuming the rate stays constant forever. Real populations level off and prices change; long projections are what the model says, not a forecast.
Before you calculate, predict the direction: a positive rate should give a bigger answer, a negative rate a smaller one. Then state the model, substitute, and write the units with your answer.
Sources
Checked November 2026.
- NOAA Global Monitoring Laboratory, Radioactive decay (carbon-14 half-life of 5,730 years)
- National Institute of Justice, What is carbon dating?
Exponential growth and decay practice quiz: 10 questions
Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.
x(t) = 2,000 × 1.04^5 ≈ 2,433.31. 2,400 adds 4% of 2,000 each year (linear growth); 2,443 uses continuous growth, 2,000 × e^0.2.
0.8^2 = 0.64, so 64% is left and the decrease is 36%. 40% wrongly adds the two 20% drops.
t = ln 2 ÷ ln 1.06 ≈ 11.90 years. 12 years is the Rule of 72 estimate.
x(t) = 1,000 × e^(0.03 × 10) = 1,000 × e^0.3 ≈ 1,349.86. 1,343.92 is the periodic model, 1,000 × 1.03^10.
r = (800 ÷ 500)^(1/6) − 1 = 1.6^(1/6) − 1 ≈ 0.0815. 10% divides the 60% total by 6; 7.83% is the continuous rate k = ln(1.6) ÷ 6.
t = ln 3 ÷ ln 1.05 ≈ 22.52 years. 21.97 uses the continuous formula ln 3 ÷ 0.05; 14.21 is only the doubling time.
x0 = 1,464.10 ÷ 1.1^4 = 1,464.10 ÷ 1.4641 = 1,000. 1,045.79 divides by 1.4 (simple growth).
Quiz complete
Frequently asked questions
How do you calculate exponential growth?
Multiply the starting amount by the growth factor raised to the time: x(t) = x0(1 + r)t. For 200 growing 3% a year for 5 years, 200 × 1.035 ≈ 231.85.
What is the difference between continuous and periodic growth?
Periodic growth adds the percentage once per step; continuous growth adds it at every instant, using ekt. For the same percentage, continuous growth ends slightly higher: 1,648.72 against 1,628.89 for 1,000 at 5% for 10 steps.
How do I model exponential decay?
Use a negative rate. A 10% decrease per period means r = −0.10, so the factor is 0.9 and x(t) = x0 × 0.9t. With a half-life T, use x(t) = x0 × (1/2)t/T.
What is doubling time?
The time an amount takes to double at a constant growth rate: ln 2 ÷ ln(1 + r). At 5% per year it is about 14.21 years. The half-life is the same idea for decay.
How do you find the growth rate from two values?
Divide the later value by the earlier one, take the t-th root and subtract 1: r = (x(t) ÷ x0)1/t − 1. Use the Rate tab to see each step.
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