Question

The number of solutions of ${\log }_4(x-1)={\log }_2(x-3)$ will be

• A. $3$
• B. $1$
• C. $2$
• D. $0$

Answer 1

Answer: (B)

$1$

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