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The curve y = a·bˣ that fits growth or decay data, with the rate per step.
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The Coddy math course teaches the method itself - you work each step on an interactive board and get told exactly where a move went wrong.
A straight line adds the same amount for every step of x. Many things multiply by the same factor instead: money at compound interest, a bacteria culture, the charge left on a battery, the temperature gap between a cooling drink and the room. That pattern is y = a·bˣ, where a is the value at x = 0 and b is the factor applied at each step. If b is above 1 the quantity grows, if b is between 0 and 1 it decays, and (b minus 1) as a percentage is the rate per step.
The standard way to fit it is to take the natural logarithm of every y. If y = a·bˣ, then ln y = ln a + x·ln b, which is a straight line in x. So the method fits an ordinary least squares line to the points (x, ln y), reads ln b off its slope and ln a off its intercept, and undoes the logarithms. This is what the ExpReg command on a TI-84 and the exponential trendline in Excel do, and this page shows every stage: the ln y column, the fitted line, and the two exponentials that bring a and b back.
Because logarithms are irrational, this is the one regression calculator in the set whose answers are rounded decimals rather than exact fractions, and it says so beside the result. It also needs every y to be above 0, since the logarithm of 0 or of a negative number does not exist. The r² shown is the r² of the straight-line fit to ln y, which is the figure a graphing calculator reports.
Replace each y with ln y. For x = 0, 1, 2, 3, 4 and y = 3, 6, 12, 24, 48 the new column is about 1.0986, 1.7918, 2.4849, 3.1781 and 3.8712.
Use the ordinary least squares formulas for the slope and intercept on the points (x, ln y). Here the slope is about 0.6931 and the intercept about 1.0986.
The slope is ln b and the intercept is ln a, so b = e to the slope and a = e to the intercept. Here b ≈ e^0.6931 = 2 and a ≈ e^1.0986 = 3.
y ≈ 3·2ˣ, or y ≈ 3e^(0.6931x) in base e. b = 2 means each step multiplies y by 2: growth of 100% per step, and a doubling time of exactly one step.
Each row is one data set typed into the calculator above. All values are rounded to four decimal places.
| x values | y values | Curve of best fit | Change per step | r² of ln y |
|---|---|---|---|---|
| 0, 1, 2, 3, 4 | 3, 6, 12, 24, 48 | y ≈ 3 · 2ˣ | +100% | 1 |
| 0, 1, 2, 3, 4, 5 | 100, 82, 67, 55, 45, 37 | y ≈ 99.9517 · 0.8195ˣ | -18.05% | 1.0000 |
| 1, 2, 3, 4, 5 | 2.1, 4.3, 8.9, 17.5, 36 | y ≈ 1.0413 · 2.0313ˣ | +103.13% | 0.9999 |
| 0, 1, 2, 3, 4 | 1000, 1050, 1102.5, 1157.625, 1215.50625 | y ≈ 1000 · 1.05ˣ | +5% | 1 |
| 0, 1, 2, 3, 4 | 5, 4, 3, 2, 1 | y ≈ 5.6968 · 0.6762ˣ | -32.38% | 0.9473 |
x: 0, 1, 2, 3, 4; y: 3, 6, 12, 24, 48Each y is twice the one before, so the points lie exactly on y = 3·2ˣ and the calculator returns a ≈ 3 and b ≈ 2, a growth of 100% per step, with r² = 1. In base e the same curve is y ≈ 3e^(0.6931x), and 0.6931 is ln 2. Carried on to x = 6 it predicts 3·2⁶ = 192.
x: 0, 1, 2, 3, 4, 5; y: 100, 82, 67, 55, 45, 37Read y as the temperature gap to the room, measured every five minutes. The fit is y ≈ 99.9517 · 0.8195ˣ, or y ≈ 99.9517e^(-0.1991x): about 18.05% of the gap disappears each step, and the half-life is ln 2 / 0.1991, about 3.48 steps. The log fit is almost perfect: r² is 0.99999 to five places, which the calculator rounds to 1.
x: 1, 2, 3, 4, 5; y: 2.1, 4.3, 8.9, 17.5, 36The values roughly double each step without doing so exactly. The fit is y ≈ 1.0413 · 2.0313ˣ, a growth of about 103.13% per step, with r² ≈ 0.9999 on the log scale. In base e, k ≈ 0.7087, so the doubling time is ln 2 / 0.7087, about 0.98 steps.
x: 0, 1, 2, 3, 4; y: 1000, 1050, 1102.5, 1157.625, 1215.50625A balance of 1000 earning 5% a year. The calculator recovers y ≈ 1000 · 1.05ˣ, a change of +5% per step, with r² = 1. Reading the interest rate back from a list of balances is exactly what this fit is for.
x: 0, 1, 2, 3, 4; y: 5, 4, 3, 2, 1These points fall by the same amount each step, so they lie on a straight line, y = -x + 5. An exponential fit still returns y ≈ 5.6968 · 0.6762ˣ with r² ≈ 0.9473 on the log scale, which looks respectable. Plot the curve against the points and it misses at both ends: a straight line is the right model here, and linear regression fits it exactly.
