Question

Solve the following equation:
$2{\log }_2\frac{x-7}{x-1}+{\log }_2\frac{x-1}{x+1}=1$

Answer 1

$2{log}_2\left(\frac{x-7}{x-1}\right)+{log}_2\left(\frac{x-1}{x+1}\right)=1\Rightarrow {\left(\frac{x-7}{x-1}\right)}^2\left(\frac{x-1}{x+1}\right)=2\Rightarrow \frac{{(x-7)}^2}{(x-1)(x+1)}=2,x\ne 1\Rightarrow x^2-14x+49={2x}^2-2\Rightarrow x^2+14x-51=0\Rightarrow x=-17,3$

But $x=3$ is not valid. Therefore, $x=-17$ is only solution.