# How do you solve e^x = 0.02?

Jul 13, 2016

$x = \ln \left(0.02\right) \approx - 3.912$

#### Explanation:

Based on our knowledge of the meaning of the natural logarithm:

Given
$\textcolor{w h i t e}{\text{XXX}} {e}^{x} = 0.02$

$\ln \left({e}^{x}\right) = \ln \left(0.02\right)$

$x = \ln \left(0.02\right)$

$x \approx - 3.912$ (using a calculator)