Question

If ${\log }_5(x+1)-1=1+{\log }_5(x-1)$, then value of $x$ is

• A. $1\frac{3}{4}$
• B. $1\frac{1}{12}$
• C. $1\frac{7}{16}$
• D. $1\frac{1}{19}$

Answer 1

The correct option is B $1\frac{1}{12}$

${\log }_5(x+1)-1=1+{\log }_5(x-1)$.
${\log }_5(x+1)-{\log }_5(x-1)=2$.
${\log }_5\frac{(x+1)}{(x-1)}=2$ ...($\because \log a=\log b=\log \frac{a}{b}$)

Converting in exponential form, we get
$5^2=\frac{(x+1)}{(x-1)}$.

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