Re : Equation involving Logs

$\begin{aligned} \log_4 (x-1) & = \log_2 (x-3) & \small \color{#3D99F6} \text{Converting the same log base on both sides} \\ \frac {\log (x-1)}{\log 4} & = \frac {\log (x-3)}{\log 2} \\ \frac {\log (x-1)}{2 \log 2} & = \frac {\log (x-3)}{\log 2} \\ \log (x-1) & = 2\log (x-3) \\ \log (x-1) & = \log (x-3)^2 \\ x-1 & = (x-3)^2 \\ x-1 & = x^2 - 6x + 9 \\ \implies x^2 - 7x + 10 & = 0 \\ (x-2)(x-5) & = 0 \\ \implies x & = \boxed 5 & \small \color{#3D99F6} \text{Since } x = 2 \implies \log_2 (x-3) \text{ is not real.} \end{aligned}$

Let LHS = RHS = t. So from LHS t^4 = (x-1) , from RHS t^2 = (x-3)

So (x-1) = (x-3)^2 = x^2 - 6x + 9 so we have a quadratic equation to solve which reads as x^2 - 7x + 10 = 0, can be factorized as (x-2_(x-5) = 0 so x = 2 or x = 5.
If x = 2 log of (x-3) does not exist as x - 3 will be less than 0 , so the only value of x which satisfies the equation in the question is when x = 5