# How do you solve 4e^(9x-1)=64?

Mar 8, 2016

$x = \frac{\ln \left(16\right) + 1}{9}$

#### Explanation:

Divide both sides of the equation first. This will deal with the $4$ attached to the ${e}^{9 x - 1}$.

${e}^{9 x - 1} = 16$

Now, take the natural logarithm of both sides of the equation.

$\ln \left({e}^{9 x - 1}\right) = \ln \left(16\right)$

Note that the $\ln$ and ${e}^{x}$ functions are inverses, so they undo one another.

$9 x - 1 = \ln \left(16\right)$

Solving through basic algebra,

$x = \frac{\ln \left(16\right) + 1}{9}$