Exponential models from context and data

In $f(x) = a \cdot b^x$, set x = 0 and the factor $b^0 = 1$ disappears: f(0) = a. So a is the initial value, full stop. The b is the per-unit factor: every time x increases by 1, the output is multiplied by b, whatever it was before.

Those jobs never trade places. The number out front is not a growth rate, and the base is not a starting amount. When a model is handed to you, read it in one breath: "starts at a, multiplies by b each step." When you build one, decide those two numbers separately and place them deliberately.

Contexts speak in percents; the model needs a factor. The bridge is b = 1 + r, with r as a decimal and negative for decay:

1. "Grows 6% per year": r = 0.06, so b = 1.06. Not 0.06, not 6.
2. "Loses 6% per year": r = −0.06, so b = 0.94. Keeping 94% is the same fact as losing 6%.
3. "Doubles": b = 2. "Loses half": b = 0.5.

Read it backward the same way: b = 1.045 means 4.5% growth per unit, because the 1 is the amount you already had. A base below 1 is decay; a base of 0.12 does not mean 12% growth, it means keeping only 12%, an 88% collapse per step.

If the data hands you f(0), that is a directly. Otherwise, two points determine the model, and division is the tool: for points $(x_1, y_1)$ and $(x_2, y_2)$,

$\dfrac{y_2}{y_1} = b^{\,x_2 - x_1}$

because the a cancels in the ratio. The exponent is the gap between the inputs, so the ratio of outputs is the factor applied over that many steps. Take the appropriate root to get the per-unit b, then substitute back for a. Example: f(1) = 6 and f(3) = 54 give $b^2 = 9$, so b = 3 and a = 6/3 = 2. Check both points before moving on; a fitted model that misses one of its own data points is arithmetic gone wrong.

An exponential function multiplies by equal factors over equal intervals. A doubling time of d units means the factor over any d-unit stretch is 2, wherever the stretch sits. A half-life of h units means the factor over any h units is one half.

To use one, count intervals, not units: 20 years with a 5-year doubling time is 20/5 = 4 doublings, a factor of $2^4 = 16$. The most reliable form writes the interval count in the exponent directly: $f(t) = a \cdot 2^{t/d}$ or $f(t) = a \cdot (0.5)^{t/h}$. The unit of the factor matters as much as the factor: a base of 2 per 5 years is not a base of 2 per year, and mixing those up changes the answer by orders of magnitude.