# How do I solve for x? 2e^{2x}-5e^x-3=0 This problem is about the natural logarithm function.

Jun 9, 2016

$x = {\log}_{e} 3$

#### Explanation:

Making ${e}^{x} = y$ and substituting we have the equivalent problem.

$2 {y}^{2} - 5 y - 3 = 0$

Solving for $y$ gives

$y = \left\{- \frac{1}{2} , 3\right\}$ we take the consistent one which is

$y = 3 = {e}^{x}$ because ${e}^{x} > 0 \forall x \in \mathbb{R}$ and solving for $x$ gives

$x = {\log}_{e} 3$