Question

The number of solutions of ${\log }_4(x-1)={\log }_2(x-3)$ is/are

• A. $3$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is B $1$

${\log }_4(x-1)={\log }_2(x-3)$, where $x>3$
$\Rightarrow \frac{\log (x-1)}{\log 4}=\frac{\log (x-3)}{\log 2},[\because {\log }_{b}a=\frac{\log a}{\log b}]$
$\Rightarrow \frac{\log (x-1)}{\log 2^2}=\frac{\log (x-3)}{\log 2}$
$\Rightarrow \frac{\log (x-1)}{2\log 2}=\frac{\log (x-3)}{\log 2}$
$\Rightarrow \log (x-1)=2\log (x-3)$
$\Rightarrow \log (x-1)=\log {(x-3)}^2,[\because x\log a=\log a^x]$
$\Rightarrow x-1={(x-3)}^2$
$\Rightarrow x^2-7x+10=0$
$\Rightarrow (x-5)(x-2)=0$
$\because x>3\therefore x=5$
Ans: B