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

May 31, 2016

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

#### Explanation:

Since ${a}^{-} b = \frac{1}{a} ^ b$, the equation can be rewritten as

${2}^{x} - \frac{1}{2} ^ x = 5$

Multiply each term by ${2}^{x}$.

${2}^{x} \left({2}^{x}\right) - \frac{1}{2} ^ x \left({2}^{x}\right) = 5 \left({2}^{x}\right)$

To simplify ${2}^{x} \left({2}^{x}\right)$, use the rule ${a}^{b} \left({a}^{c}\right) = {a}^{b + c}$:

${2}^{x + x} - 1 = 5 \left({2}^{x}\right)$

${2}^{2 x} - 5 \left({2}^{x}\right) - 1 = 0$

Let $u = {2}^{x}$. Then, ${2}^{2 x} = {\left({2}^{x}\right)}^{2} = {u}^{2}$ and $5 \left({2}^{x}\right) = 5 u$.

${u}^{2} - 5 u - 1 = 0$

Use the quadratic formula to show that

$u = \frac{- \left(- 5\right) \pm \sqrt{{5}^{2} - \left(4 \cdot - 1 \cdot 1\right)}}{2 \cdot 1} = \frac{5 \pm \sqrt{29}}{2}$

Recall that $u = {2}^{x}$.

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

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

Using ${\log}_{a} \left(\frac{b}{c}\right) = {\log}_{a} \left(b\right) - {\log}_{a} \left(c\right)$:

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

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

As for the negative version of $\frac{5 \pm \sqrt{29}}{2}$, we see that

${2}^{x} = \frac{5 - \sqrt{29}}{2}$

Has no solutions since $\frac{5 - \sqrt{29}}{2} < 0$ and ${2}^{x} > 0$ for all values of $x$.

May 31, 2016

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

#### Explanation:

Taking $y = {2}^{x}$ we have

$y - \frac{1}{y} = 5 \to {y}^{2} - 5 y - 1 = 0$

Solving for $y$

$y = \frac{1}{2} \left(5 - \sqrt{29}\right)$, $y = \frac{1}{2} \left(5 + \sqrt{29}\right)$

but $\frac{1}{2} \left(5 - \sqrt{29}\right) < 0$ and $y = {2}^{x} > 0$ Now solving for $x$

so

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