Question:
The number of solutions of the equation $\log _{4}(x-1)=\log _{2}(x-3)$ is
Solution:
$\frac{1}{2} \log _{2}(x-1)=\log _{2}(x-3)$
$x-1=(x-3)^{2}$
$x^{2}-6 x+9=x-1$
$x^{2}-7 x+10=0$
$x=2,5$
$X=2 \quad$ Not possible as $\log _{2}(x-3)$ is not defined
$\Rightarrow$ No. of solution $=1$
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