Question

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

• A. $3$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is B

$1$

Explanation for correct option

Given, ${\log }_4\left(x-1\right)={\log }_2\left(x-3\right)$

$$\begin{gathered} \Rightarrow {\log }_4\left(x-1\right)={\log }_{4^{\frac{1}{2}}}\left(x-3\right) \\ \Rightarrow {\log }_4\left(x-1\right)=2{\log }_4\left(x-3\right) \\ \Rightarrow {\log }_4\left(x-1\right)={\log }_4{\left(x-3\right)}^2 \\ \Rightarrow x-1={\left(x-3\right)}^2 \\ \Rightarrow x-1=x^2-6x+9 \\ \Rightarrow x^2-7x+10=0 \\ \Rightarrow x^2-5x-2x+10=0 \\ \Rightarrow x\left(x-5\right)-2\left(x-5\right)=0 \\ \Rightarrow \left(x-5\right)\left(x-2\right)=0 \\ \therefore x=2,5 \end{gathered}$$

At $x=2,\log \left(x-3\right)$ is undefined.

At, $x=5,\log \left(x-3\right)$ will be $\log 2$,

$\therefore x=5$

So, one solution exist

Hence, the correct option is $\mathrm{Option}\left(\mathrm{B}\right)$