quadratic 10

We'll use some basic properties of logarithms that one can review from the Logarithms Brilliant wiki (if they aren't familiar with logarithms).

---

What is given to us is this:

$log_4(x-1)=\log_2(x-3)$

We will proceed assuming that $(x-1),(x-3)\gt 0$ because otherwise the equation will be undefined on one or both sides. So,

$\log_4(x-1)=\log_2(x-3)\\ \implies \frac{\ln (x-1)}{\ln 4}=\frac{\ln (x-3)}{\ln 2}\\ \implies \frac{\ln (x-1)}{2}=\ln (x-3)\\ \implies \ln \left(\frac{x-1}{(x-3)^2}\right)=0\\ \implies \frac{x-1}{(x-3)^2}=1$

This step that we just did may or may not have introduced extraneous solutions that aren't solutions to the original equation. This is because of the assumption we made at the beginning. Let's proceed further. The equation becomes,

$x^2-7x+10=0 \implies (x-5)(x-2)=0\implies x=5,2$

Now, we will check if any one of them is an extraneous solution that doesn't satisfy the original equation by matching them against the assumptions made at the beginning so as to achieve the solutions for which the original equation doesn't become undefined and satisfies it.

$(5-1)=4\gt 0~,~(5-3)=2\gt 0~,~(2-1)=1\gt 0~,~(2-3)=(-1) \not\gt 0$

So, $x=3$ is not a solution as it makes RHS of original equation undefined. Thus, we have the only possible solution $x=5$ . Hence, the original equation has only one solution.