Question

Find the value of $x$ which satisfies the equation: ${\log }_{\frac{1}{5}}\frac{x+2}{10}+{\log }_5\left(\frac{2}{x+1}\right)=0$

Answer 1

${\log }_{\frac{1}{5}}\frac{x+2}{10}+{\log }_5\left(\frac{2}{x+1}\right)=0$ is valid when $x+2>0,x+1>0$
$\Rightarrow x>-1$
Changing base, we can write
$-{\log }_5\frac{x+2}{10}+{\log }_5\left(\frac{2}{x+1}\right)=0,[\because {\log }_{\frac{1}{x}}y=-{\log }_{x}y]$
$\Rightarrow {\log }_5\frac{x+2}{10}={\log }_5\left(\frac{2}{x+1}\right)$
$\Rightarrow \frac{x+2}{10}=\frac{2}{x+1}\Rightarrow x=-6,3$
As $x>-1,\therefore x=3$ is the solution
Ans: 3