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

Jul 8, 2016

$x = {\log}_{2} \left(5 + \sqrt{29}\right) - 1 \approx 2.3764522$

#### Explanation:

Let $t = {2}^{x}$

Then the equation becomes:

$t - \frac{1}{t} = 5$

Multiply through by $4 t$ and rearrange slightly to get:

$0 = 4 {t}^{2} - 20 t - 4$

$= {\left(2 t - 5\right)}^{2} - 29$

$= {\left(2 t - 5\right)}^{2} - {\left(\sqrt{29}\right)}^{2}$

$= \left(\left(2 t - 5\right) - \sqrt{29}\right) \left(\left(2 t - 5\right) + \sqrt{29}\right)$

$= \left(2 t - 5 - \sqrt{29}\right) \left(2 t - 5 + \sqrt{29}\right)$

Hence:

${2}^{x + 1} = 2 \cdot {2}^{x} = 2 t = 5 \pm \sqrt{29}$

For any Real value of $a$, ${2}^{a} > 0$.

So (for Real solutions) we require:

${2}^{x + 1} = 5 + \sqrt{29}$

Hence:

$x + 1 = {\log}_{2} \left(5 + \sqrt{29}\right)$

So:

$x = {\log}_{2} \left(5 + \sqrt{29}\right) - 1$

To calculate a numerical value for ${\log}_{2} \left(5 + \sqrt{29}\right)$ we can use the change of base formula to find:

${\log}_{2} \left(5 + \sqrt{29}\right) = \ln \frac{5 + \sqrt{29}}{\ln} \left(2\right) \approx 3.3764522$

So:

$x \approx 3.3764522 - 1 = 2.3764522$