# Solve the following equation...? 2^(4x) - 5(2^(2x - 1/2)) + 2 = 0

Nov 26, 2017

$x = \ln \frac{\frac{25 \pm \sqrt{609}}{2 \sqrt{2}}}{\ln 4}$

#### Explanation:

${2}^{4 x} - 5 \left({2}^{2 x - \frac{1}{2}}\right) + 2 = 0 \iff$

${2}^{{\left(2 x\right)}^{2}} - 5 \cdot {2}^{2 x} \textcolor{red}{\times} 5 \cdot {2}^{- \frac{1}{2}} + 2 = 0 \iff$

${\left({2}^{2 x}\right)}^{2} - \left(\frac{25}{\sqrt{2}}\right) {2}^{2 x} + 2 = 0 \iff$

Now the quadratic equation should be easy to see.
You have to replace ${2}^{2 x}$ with an y.

<=> y^2−(25/(√2))y+2=0

$y = \frac{\frac{25}{\sqrt{2}} \pm \sqrt{\frac{625}{2} - 2 \cdot 2 \cdot 2}}{2}$

$y = \frac{\frac{25}{\sqrt{2}} \pm \sqrt{\frac{609}{2}}}{2}$

${2}^{2 x} = y = \frac{\frac{25}{\sqrt{2}} \pm \sqrt{\frac{609}{2}}}{2}$

Appyling logarithms:

$2 x \ln 2 = \ln \left(\frac{25 \pm \sqrt{609}}{2 \sqrt{2}}\right)$

$x = \ln \frac{\frac{25 \pm \sqrt{609}}{2 \sqrt{2}}}{2 \ln 2}$

$x = \ln \frac{\frac{25 \pm \sqrt{609}}{2 \sqrt{2}}}{\ln 4}$