Question

Solve the following equation:
$({\log }_{4}x-2)\cdot {\log }_{4}x=\frac{3}{2}({\log }_{4}x-1)$

Answer 1

${(\log }_{4}x-2\left)\right({\log }_{4}x)=\frac{3}{2}({\log }_{4}x-1)\Rightarrow {{(\log }_{4}x)}^2-2{\log }_{4}x=\frac{3}{2}{\log }_{4}x-\frac{3}{2}\Rightarrow {{(\log }_{4}x)}^2-\frac{7}{2}{(\log }_{4}x)+\frac{3}{2}=0\Rightarrow {\log }_{4}x=3,\frac{1}{2}\Rightarrow x=64,2.$