Question

If $a^x=N$ then ${\log }_{a}N=x$ and ${\log }_{2}ab={\log }_{2}a+{\log }_{2}b$, then what are the number of solutions of the equation ${\log }_4(x-1)={\log }_2(x-3)$?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\frac{\log (x-1)}{\log 4}=\frac{\log (x-3)}{\log 2}$

$\Rightarrow \frac{\log (x-1)}{2}=\log (x-3)$

$\Rightarrow \log (x-1)=2\left[\log \right(x-3\left)\right]$

$\Rightarrow \log (x-1)=\log (x-3{)}^2$

$\Rightarrow (x-1)=(x-3{)}^2$

$\Rightarrow x^2-7x+10=0$

$\Rightarrow x=2$, $x=5$

$\Rightarrow x=5$ [since for $x=2$, $\log (x-3)$ is not defined]

$\therefore$ Number of solutions $=1$