Question

Solve for $x$

$2{\log }_2\frac{x-7}{x-1}+{\log }_2\frac{x-1}{x+1}=1$

Answer 1

$2{\log }_2\frac{x-7}{x-1}+{\log }_2\frac{x-1}{x+1}=1$

$\Rightarrow {\log }_2{\left(\frac{x-7}{x-1}\right)}^2+{\log }_2\frac{x-1}{x+1}=1$

$\Rightarrow {\left(\frac{x-7}{x-1}\right)}^2\times \frac{x-1}{x+1}=2^1=2$

$\Rightarrow \frac{x^2-14x+49}{x^2-1}=2$

$\Rightarrow x^2-14x+49=2x^2-2$

$\Rightarrow x^2+14x-51=0$

$\Rightarrow x^2+17x-3x-51=0$

$\Rightarrow x(x+17)-3(x+17)=0$

$\Rightarrow x-3=0,x+17=0$

$\therefore x=3,-17$

Substitute $x=-17$ in $2{\log }_2\frac{x-7}{x-1}+{\log }_2\frac{x-1}{x+1}=1$ we get

$2{\log }_2\frac{-17-7}{-17-1}+{\log }_2\frac{-17-1}{-17+1}$

$=2{\log }_2\frac{-24}{-18}+{\log }_2\frac{-18}{-16}$

$={\log }_2\frac{{24}^2}{{18}^2}\times {\log }_2\frac{18}{16}$

$={\log }_2(\frac{24}{18}\times \frac{24}{18}\times \frac{18}{16})$

$={\log }_{2}2=1$

Hence $x=-17$

$x=3$ is not valid solution

$\because$ for $x=3$ the $\log$ is not defined.