Question

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

Answer 1

`x=log(5+-sqrt 21)/log 2 -1 = +-2.2604.`, nearly.

Explanation:

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

So, `y = 2^x = (5+-sqrt21)/2.` Both are positive and , therefore,

admissible. Equating logarithms and rearranging,

`x=log(5+-sqrt 21)/log 2 -1=+-2.2604`, nearly

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