To write an exponential model you need a starting value and a factor. A story may give both, as in growth and decay rates, or you may have to find them from data points, as with the ratio of a geometric sequence.
Why it works
In $y = a \cdot b^x$, the point at $x = 0$ gives $a$ directly, because $b^0 = 1$. A second point then pins down $b$: if $y$ goes from $a$ at $x = 0$ to $y_2$ at $x = 2$, then $a \cdot b^2 = y_2$, so $b^2 = \frac{y_2}{a}$. In data with equal steps, the factor shows up as the constant ratio of neighboring values. If the differences are constant instead, a line fits better.
Key idea
Find $a$ (the value at $x = 0$) and $b$ (the factor per step, from a ratio of outputs), then write $y = a \cdot b^x$. Check that the model reproduces the data.
Worked examples
Worked example: From two points
An exponential function passes through $(0, 5)$ and $(2, 45)$. Write it.
- $x = 0$ gives $a = 5$.
- $5 \cdot b^2 = 45$, so $b^2 = 9$ and $b = 3$ ($b$ must be positive).
- Check: $5 \cdot 3^1 = 15$ would be the value at $x = 1$, and $5 \cdot 3^2 = 45$.
Answer: $y = 5 \cdot 3^x$
Worked example: From a story
A sample of $400$ mg of a substance loses half its mass every $6$ hours. Write a model.
- Every $6$ hours the mass is multiplied by $\frac{1}{2}$.
- After $t$ hours there have been $\frac{t}{6}$ half-lives.
Answer: $M(t) = 400\left(\frac{1}{2}\right)^{t/6}$
Worked example: Choose the model
Data: $(0, 4), (1, 10), (2, 16), (3, 22)$. Linear or exponential?
- Differences: $6, 6, 6$, constant. Ratios: $2.5, 1.6, 1.375$, not constant.
Answer: Linear: $y = 6x + 4$.
Common mistake
Fitting a linear model to data that changes by a constant ratio. Data like $5, 15, 45, 135$ has differences $10, 30, 90$ that keep growing, so a straight line will miss badly after a few steps. Always compute both differences and ratios before choosing.