Question

Solve the following equation:
${\log }_2\left(\frac{x}{4}\right)=\frac{15}{{\log }_2\frac{x}{8}-1}$

Answer 1

${log}_2\left(\frac{x}{4}\right)=\frac{15}{{log}_2\left(\frac{x}{8}\right)-1}\Rightarrow {log}_{2}x-2=\frac{15}{{log}_{2}x-4}\Rightarrow {(log}_{2}x-2\left)\right({log}_{2}x-4)=15\Rightarrow {(log}_{2}x{)}^2-{6log}_{2}x+8=15\Rightarrow {{(log}_{2}x)}^2-{6log}_{2}x-7=0$

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

$x=2^{7,}{2}^{-1}$

Therefore, $x=128,\frac{1}{2}.$