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The number of solutions of the equation $\log_4(x-1) = \log_2(x-3)$ is
$\frac{1}{2}\log_2(x-1) = \log_2(x-3)$
$x - 1 = (x-3)^2$
$x^2 - 6x + 9 = x - 1$
$x^2 - 7x + 10 = 0$
$x = 2, 5$
$X = 2$: Not possible as $\log_2(x-3)$ is not defined
$\Rightarrow$ No. of solution $= 1$
BETA
