Question

If ${\log }_5(x+1)-1=1+{\log }_5(x-1)$, then $x$ = $m\frac{1}{12}$.

Therefore, value of $m$ is equal to

• A. $-5$
• B. $2$
• C. $3$
• D. $1$

Answer 1

The correct option is D $1$

${\log }_5(x+1)-1=1+{\log }_5(x-1)$
${\log }_5(x+1)=2{\log }_{5}5+{\log }_5(x-1)$
${\log }_5(x+1)={\log }_{5}25+{\log }_5(x-1)$
${\log }_5(x+1)={\log }_{5}25(x-1)$
$\text{Removing log}$

$(x+1)=25(x-1)$
$x+1=25x-1$
$24x=26$
$x=\frac{13}{12}\Rightarrow 1\frac{1}{12}$

Value of $m$ is $1$.