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

Feb 17, 2016

$x = {\log}_{2} \left(\frac{5}{2} + \frac{\sqrt{29}}{2}\right) \approx 2.37645$

#### Explanation:

We have:

${2}^{x} - {2}^{-} x = 5$

Take the $5$ over to the other side to get:
${2}^{x} - 5 - {2}^{-} x = 0$

Now multiply the whole thing through by ${2}^{x}$ and we get:

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

We substitute $t = {x}^{2}$ and that gives us:

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

We now have a quadratic that can be solved using the quadratic formula with terms $a = 1 , b = 5$ and $c = - 1$.

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

Thus $t = \frac{5}{2} + \frac{\sqrt{29}}{2}$ or $t = \frac{5}{2} - \frac{\sqrt{29}}{2}$

Remeber ${2}^{x} = t$. Reverse the substitution of $t$ to obtain:

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

Now take the logarithm to the base $2$ and we get:

$x = {\log}_{2} \left(\frac{5}{2} + \frac{\sqrt{29}}{2}\right) \approx 2.37645$

We can ignore the $t = \frac{5}{2} - \frac{\sqrt{29}}{2}$ as it would involve taking the logarithm of a negative number which we wish to avoid.