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

Jul 14, 2015

Let $f \left(x\right) = a {b}^{x}$ where $a \ne 0$ and $b > 0$.

Use known values $f \left(1\right) = 6$ and $f \left(3\right) = 24$ and solve to find $a = 3$, $b = 2$

Hence $f \left(x\right) = 3 \cdot {2}^{x}$

#### Explanation:

The standard formula for an exponential function is:

$f \left(x\right) = a {b}^{x}$, with $a \ne 0$ and $b > 0$

Given: $f \left(1\right) = 6$ and $f \left(3\right) = 24$

$4 = \frac{24}{6} = f \frac{3}{f} \left(1\right) = \frac{a {b}^{3}}{a {b}^{1}} = {b}^{2}$

So $b = \pm \sqrt{4} = \pm 2$.

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

Then $a = \frac{a {b}^{1}}{b} = f \frac{1}{b} = \frac{6}{2} = 3$

So $a = 3$ and $b = 2$, giving us $f \left(x\right) = 3 \cdot {2}^{x}$

General case:

Given $f \left({x}_{1}\right) = {y}_{1}$ and $f \left({x}_{2}\right) = {y}_{2}$

${b}^{{x}_{2} - {x}_{1}} = \frac{a {b}^{{x}_{2}}}{a {b}^{{x}_{1}}} = f \frac{{x}_{2}}{f} \left({x}_{1}\right)$

So $b = {\left(f \frac{{x}_{2}}{f} \left({x}_{1}\right)\right)}^{\frac{1}{{x}_{2} - {x}_{1}}}$

Then $a = f \frac{{x}_{1}}{{b}^{{x}_{1}}}$