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

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Answer 1 · 2015-07-14T07:22:24

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 $f(x) = 3*2^x$

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 $a = f(x_1) / (b^(x_1))$