# How do you solve for x?: e^(-2x)+e^x = 1/3?

Oct 22, 2015

I found: $- \ln \left[3 \left({e}^{-} 2 + 1\right)\right] = - \ln \left(3\right) - \ln \left({e}^{-} 2 + 1\right)$

#### Explanation:

Collect ${e}^{x}$ on the left:
${e}^{x} \left({e}^{-} 2 + 1\right) = \frac{1}{3}$ rearrange:
${e}^{x} = \frac{1}{3 \left({e}^{-} 2 + 1\right)}$
apply the $\ln$ on both sides:
$\ln \left[{e}^{x}\right] = \ln \left[\frac{1}{3 \left({e}^{-} 2 + 1\right)}\right]$
so:
$x = \ln \left[\frac{1}{3 \left({e}^{-} 2 + 1\right)}\right]$
use the properties of log relating sums and subtractions to multiplications and divisions to get:
$x = \ln \left(1\right) - \ln \left[3 \left({e}^{-} 2 + 1\right)\right] = 0 - \ln \left(3\right) - \ln \left({e}^{-} 2 + 1\right)$