Question

Number of solutions of ${\log }_4(x-1)={\log }_2(x-3)$ is

Answer 1

${\log }_4(x-1)={\log }_2(x-3)$

$\Rightarrow {\log }_{2^2}(x-1)={\log }_2(x-3)$

$\Rightarrow \frac{1}{2}{\log }_2(x-1)={\log }_2(x-3)$

$\Rightarrow {\log }_2(x-1)=2{\log }_2(x-3)$

$\Rightarrow {\log }_2(x-1)={\log }_2{(x-3)}^2$

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

$\Rightarrow x-1=x^2-6x+9$

$\Rightarrow x^2-6x+9-x+1=0$

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

$\Rightarrow x^2-2x-5x+10=0$

$\Rightarrow x(x-2)-5(x-2)=0$

$\Rightarrow (x-2)(x-5)=0$

$\therefore x=2,5$

But for $x=2,{\log }_2(x-3)$ is negative.

Hence $x=2$ is not valid

$\therefore x=5$

Hence number of solutions is $1$