# How do I express (2^x)-1=2/(2^x) as a quadratic equation, and show that only one real solution, x=1, exists?

Feb 21, 2018

See explanation...

#### Explanation:

Given:

${2}^{x} - 1 = \frac{2}{{2}^{x}}$

Multiply both sides by ${2}^{x}$ to get:

${\left({2}^{x}\right)}^{2} - \left({2}^{x}\right) = 2$

This is a quadratic equation in ${2}^{x}$. If you wish you can use a substitution $t = {2}^{x}$ to get ${t}^{2} - t = 2$, but I will just leave it as ${2}^{x}$...

Subtract $2$ from both sides to get:

$0 = {\left({2}^{x}\right)}^{2} - \left({2}^{x}\right) - 2 = \left({2}^{x} - 2\right) \left({2}^{x} + 1\right)$

So:

${2}^{x} = 2 \text{ }$ or $\text{ } {2}^{x} = - 1$

Note that for any real value of $x$ we have ${2}^{x} > 0$

Hence the only possible real solutions are given by:

${2}^{x} = 2$

The function $f \left(x\right) = {2}^{x}$ is strictly monotonically increasing and therefore one to one.

We find:

${2}^{\textcolor{red}{1}} = 2$

So $x = 1$ is the unique real solution.