# How do you solve e^x+e^-x=4?

Aug 22, 2015

$x = \ln \left(2 + \sqrt{3}\right) \mathmr{and} x = \ln \left(2 - \sqrt{3}\right)$

#### Explanation:

Temporarily simplify by letting $y = {e}^{x}$

${e}^{x} + {e}^{- x} = 4$
becomes $y + \frac{1}{y} = 4$

Multiplying by $y$ and sifting things around:
$\textcolor{w h i t e}{\text{XXXX}} {y}^{2} - 4 y + 1 = 0$
Applying the quadratic formula, we get
$\textcolor{w h i t e}{\text{XXXX")y =2+sqrt(3)color(white)("XXXX")orcolor(white)("XXXX}} y = 2 - \sqrt{3}$

If $y = {e}^{x} = 2 + \sqrt{3}$
$\textcolor{w h i t e}{\text{XXXX}} \ln \left({e}^{x}\right) = x = \ln \left(2 + \sqrt{3}\right)$

Similarly, if $y = {e}^{x} = 2 - \sqrt{3}$
$\textcolor{w h i t e}{\text{XXXX}} x = \ln \left(2 - \sqrt{3}\right)$