# How do you solve e^x>30?

Jul 8, 2018

$x > \ln \left(30\right)$

#### Explanation:

It may be helpful to look at a graph of ${e}^{x}$. Notice that because it is an exponential function with a base greater than 1, it is monotonically increasing, so:

${e}^{a} > {e}^{b} \implies a > b$

To clarify, I'll use a concrete example.

${e}^{2} > {e}^{0} \implies 2 > 0$

Getting back to our original problem, we want to find when ${e}^{x}$ is greater than $30$. Rewrite $30$ as ${e}^{\ln} \left(30\right)$

${e}^{x} > {e}^{\ln} \left(30\right)$

But if ${e}^{x} > {e}^{\ln} \left(30\right)$:

$x > \ln \left(30\right)$

Technically, you could also take the natural logarithm on both sides since that is also monotonically increasing.