Find the value of $x$ satisfying $\log_3(x^2 - 4x + 3) = \log_3(x - 1)$.

First, find the domain constraints for the logarithms to be defined:

1. $x^2 - 4x + 3 > 0 \implies (x-1)(x-3) > 0 \implies x \in (-\infty, 1) \cup (3, \infty)$
2. $x - 1 > 0 \implies x > 1$

Intersecting these two conditions gives the domain: $x > 3$.

Now equate the arguments:

$$x^2 - 4x + 3 = x - 1$$

$$x^2 - 5x + 4 = 0$$

$$(x - 1)(x - 4) = 0 \implies x = 1 or x = 4$$

Comparing with our domain condition ($x > 3$), $x = 1$ is excluded.
Therefore, $x = 4$.

Answer: 4