Exponential Fit Calculator (2 Points)
Fit y = A·e^(kx) through two points, recovering the amplitude A and growth-rate constant k, then predict y at any x.
How it works
k = ln(y₂ / y₁) / (x₂ − x₁) · A = y₁ / e^(k·x₁) · per-unit rate = e^k − 1
Two positive data points determine a unique exponential curve. Taking the log of the ratio of the y-values linearises the problem: the slope of ln(y) versus x is the growth constant k, and back-substituting one point recovers the amplitude A. A positive k means growth and a negative k means decay; converting to e^k − 1 expresses it as a familiar percentage rate per unit of x.
Worked example
Through (0, 2) and (3, 16): k = ln(16/2)/3 = ln 8 / 3 ≈ 0.6931, and A = 2 since x₁ = 0. The curve y = 2·e^(0.6931x) doubles the exponent each time x rises by 1 (e^0.6931 = 2), so at x = 5 it predicts 2 × 2⁵ = 64.
Sources: Exponential growth — Wikipedia
Frequently asked questions
Why must both y-values be positive?
The model y = A·e^(kx) with a positive A is always positive and never crosses zero, so it cannot pass through a point with a zero or negative y. If your data changes sign, an exponential is the wrong model.
What does the growth constant k mean?
k is the continuous growth rate. Positive k means the quantity grows exponentially, negative k means it decays. Converting with e^k − 1 turns it into the percentage change per unit of x, which is often easier to read.
How does this relate to doubling time?
For growth, the doubling time is ln(2)/k in the same units as x. In the example k ≈ 0.6931, so the doubling time is exactly 1 — consistent with the value doubling every time x increases by 1.
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This calculator is for educational purposes. Double-check important results before acting on them.