# How do you solve  2^x+2^(-x)=5?

Jul 26, 2016

#### Answer:

$x = \log \frac{5 \pm \sqrt{21}}{\log} 2 - 1 = \pm 2.2604 .$, nearly.

#### Explanation:

This is a quadratic equation ${y}^{2} - 5 y + 1 = 0$, where $y = {2}^{x}$.

So, $y = {2}^{x} = \frac{5 \pm \sqrt{21}}{2.}$ Both are positive and , therefore,

admissible. Equating logarithms and rearranging,

$x = \log \frac{5 \pm \sqrt{21}}{\log} 2 - 1 = \pm 2.2604$, nearly

The given equation has an alternative form cosh(x log 2) = 2.5.