Question

Solve the following equation.
${\log }_{1/2}^2\left(4x\right)+{\log }_2\left(\frac{x^2}{8}\right)=8$

Answer 1

${log}_{\frac{1}{2}}^2\left(4x\right)+{log}_2\left(\frac{x^2}{8}\right)=8\Rightarrow (-{log}_{2}4x{)}^2+{log}_2{x}^2-lo{g}_{2}8=8\Rightarrow {(2+{log}_{2}x)}^2+{2log}_{2}x-3=8$

Let, ${log}_{2}x=a$

Therefore,

$({2+a)}^2+2a=11\Rightarrow a^2+6a-7=0\Rightarrow a=-7,1$

Therefore, ${log}_{2}x=-7,1$

$x=2^{-7},2.$