Question

How do I find the equation of an exponential function that passes through (1,6) and (3,24)?

Answer 1

Let `f(x) = ab^x` where `a != 0` and `b > 0`.

Use known values `f(1) = 6` and `f(3) = 24` and solve to find `a=3`, `b=2`

Hence `f(x) = 3*2^x`

Explanation:

The standard formula for an exponential function is:

`f(x) = ab^x`, with `a != 0` and `b > 0`

Given: `f(1) = 6` and `f(3) = 24`

`4 = 24/6 = f(3) / f(1) = (ab^3)/(ab^1) = b^2`

So `b = +-sqrt(4) = +-2`.

We want `b > 0`, so pick `b = 2`.

Then `a = (ab^1)/b = f(1)/b = 6 / 2 = 3`

So `a=3` and `b=2`, giving us `f(x) = 3 * 2^x`

General case:

Given `f(x_1) = y_1` and `f(x_2) = y_2`

`b^(x_2 - x_1) = (ab^(x_2)) / (ab^(x_1)) = f(x_2)/f(x_1)`

So `b = (f(x_2) / f(x_1))^(1/(x_2 - x_1))`

Then `a = f(x_1) / (b^(x_1))`