How many solutions are possible for $\log_4 (x-1) = \log_2 (x-3)$?

Question

$$\log_4 (x-1) = \log_2 (x-3)$$ I can calculate value of $x$ by two methods, one gives single solution whereas other one results in two values of $x$. I am confused which one in true.

Answer 1

Do not forget about the domain restrictions.

Both of your solutions must satisy $(x-3)>0$ and $(x-1)>0$.

That is, $x>3$ is must.

Answer 2

One way is to deduce that $x-1=(x-3)^2$, and solve a quadratic. However, this quadratic isn't actually equivalent to the original equation, but to the equation $$\log_4(x-1)=\log_2|x-3|.$$ One solution to the quadratic is the solution of the original equation, but the other is the solution of $$\log_4(x-1)=\log_2(3-x).$$