Question

Solve the following equation:
${\log }_2^{2}x+2{\log }_2\sqrt{x}-2=0$

Answer 1

${\log }_2^{3}x+2{\log }_2\sqrt{x}-2=0\Rightarrow {(\log }_2^x{)}^2+\left({\log }_2^x\right)-2=0\Rightarrow ({\log }_2^x{)}^2+{(\log }_3^x)-2=0\Rightarrow {(\log }_2^x+2\left)\right({\log }_2^x-1)=0\Rightarrow {\log }_2^x=-2,{\log }_2^x=1\Rightarrow x=2^{-2},x=2^1.\Rightarrow x=\frac{1}{4},2.$