Question

Solve $\sqrt{{\log }_2{x}^4}+4{\log }_4\sqrt{\frac{2}{x}}=2$

Answer 1

$\sqrt{{\log }_2{x}^4}+4{\log }_4\sqrt{\frac{2}{x}}=2$

$\Rightarrow \sqrt{4{\log }_{2}x}+4{\log }_4\sqrt{2}-4{\log }_4\sqrt{x}=2$

$\Rightarrow 2\sqrt{{\log }_{2}x}+4{\log }_4\sqrt{2}-4{\log }_4\sqrt{x}=2$

$\Rightarrow \sqrt{{\log }_{2}x}+2{\log }_4\sqrt{2}-2{\log }_4\sqrt{x}=1$

$\Rightarrow \sqrt{{\log }_{2}x}+\frac{1}{2}{\log }_{2}2-\frac{2}{2}{\log }_{2}x=1$

$\Rightarrow \sqrt{{\log }_{2}x}+\frac{1}{2}-\frac{2}{2}{\log }_{2}x=1$

$\Rightarrow \sqrt{{\log }_{2}x}-\frac{2}{2}{\log }_{2}x=1-\frac{1}{2}$

$\Rightarrow \sqrt{{\log }_{2}x}-{\log }_{2}x=\frac{1}{2}$

Let $\alpha =\sqrt{{\log }_{2}x}\Rightarrow {\alpha }^2={\log }_{2}x$

$\Rightarrow \alpha -{\alpha }^2=\frac{1}{2}$

$\Rightarrow 2\alpha -2{\alpha }^2-1=0$

$\Rightarrow 2{\alpha }^2-2\alpha -1=0$

$\Rightarrow \alpha =\frac{2\pm \sqrt{4+8}}{4}=\frac{2\pm 2\sqrt{3}}{4}=\frac{\pm \sqrt{3}}{2}$

$\Rightarrow \sqrt{{\log }_{2}x}=\frac{1+\sqrt{3}}{2}\because \sqrt{x}>0$

$\Rightarrow {\log }_{2}x={\left(\frac{1+\sqrt{3}}{2}\right)}^2=\frac{1+3+2\sqrt{3}}{4}=\frac{2+\sqrt{3}}{2}$

$\therefore x=2^{\frac{2+\sqrt{3}}{2}}$