The number of solutions

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

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

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

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

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

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

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

$\Rightarrow x=2,5$

But $x \neq 2$ because it is not satisfying the domain of given equation i.e $\log _{2}(x-3) \rightarrow$ its domain $x$ $>3$

finally $x$ is 5

$\therefore$ No. of solutions $=1$.