Question

Find the value of x satisfying the equation ${\log }_{\frac{1}{2}}(x-1)+{\log }_{\frac{1}{2}}(x+1)-{\log }_{\frac{1}{\sqrt{2}}}(7-x)=1$

Answer 1

${log}_{\frac{1}{2}}(x-1)+{log}_{\frac{1}{2}}(x+1)-{log}_{\frac{1}{\sqrt{2}}}(7-x)=1\Rightarrow {log}_{\frac{1}{2}}(x-1)(x+1)-{log}_{\frac{1}{2}}{(7-x)}^2=1\Rightarrow \frac{(x-1)(x+1)}{{(7-x)}^2}=\frac{1}{2}\Rightarrow {2x}^2-2=x^2-14x+49\Rightarrow x^2+14x-51=0\Rightarrow (x+17)(x-3)=0$

$\Rightarrow x=-17$ or $3$

But $x\ne -17$ (log cannot take negative.)

Therefore, $x=3.$