An algebra problem by A Former Brilliant Member

Relevant wiki: Rules of Exponents

$\large \dfrac{\sqrt{\dfrac{x^4}{\sqrt{64}}}}{\sqrt{\dfrac{\sqrt{x^6}}{5^2}}}=2$

Multiply both sides by $\sqrt{\dfrac{\sqrt{x^6}}{5^2}}$ . We get,

$\sqrt{\dfrac{x^4}{8}}=2\left(\sqrt{\dfrac{\sqrt{x^6}}{25}}\right)$

Square both sides

$\left(\sqrt{\dfrac{x^4}{8}}\right)^2=\left[2\left(\sqrt{\dfrac{\sqrt{x^6}}{25}}\right)\right]^2$

$\dfrac{x^4}{8}=4\left(\dfrac{\sqrt{x^6}}{25}\right)$

Square both sides again, we get

$\dfrac{x^8}{64}=16\left(\dfrac{x^6}{625}\right)$

Simplifying, we get

$\dfrac{x^8}{x^6}=\dfrac{64(16)}{625}$

$x^2=\dfrac{1024}{625}$

$x=\pm \sqrt{\dfrac{1024}{625}}$

$x=\pm \dfrac{32}{25}$

$answer=\dfrac{32}{25}=$ $\boxed{1.28}$

$\begin{aligned} \frac {\sqrt{\frac{x^4}{\sqrt{64}}}}{\sqrt{\frac{\sqrt{x^6}}{5^2}}} & = 2 \\ \sqrt{\frac{x^4}{\sqrt{64}} \times \frac {5^2}{\sqrt{x^6}}} & = 2 & \small \color{#3D99F6} \text{Squaring both sides} \\ \frac{x^4}{8} \times \frac {25}{x^3} & = 4 \\ \implies x & = \frac {4 \times 8}{25} \\ & = \boxed{1.28} \end{aligned}$