# How do you solve 7e^{x} = 9- e^{- x}?

Nov 13, 2016

Two solutions
${x}_{1} = \ln \left(\frac{9 - \sqrt[2]{53}}{14}\right)$
${x}_{2} = \ln \left(\frac{9 + \sqrt[2]{53}}{14}\right)$

#### Explanation:

Let's rewrite the equation
$7 {e}^{x} - 9 + \frac{1}{e} ^ x = 0$
We can multiply both sides for ${e}^{x}$ so that we get

$7 {e}^{2 x} - 9 {e}^{x} + 1 = 0$
this can be solved as an ordinary second degree eqaution

${e}^{x} = \frac{9 \pm \sqrt[2]{81 - 28}}{14}$

and consequently

$x = \ln \left(\frac{9 \pm \sqrt[2]{53}}{14}\right)$