First, we can write the right side of the equation as a single logarithm using the property $\log_b a + \log_b c = \log_b(ac)$: $\log_5 (2x - 10) + 1 = \log_5 (2x - 10) + \log_5 5 = \log_5 [5(2x - 10)] = \log_5 (10x - 50)$.

Now, equate the arguments of the logarithms on both sides: $x^2 - 4x - 5 = 10x - 50$ $x^2 - 14x + 45 = 0$ $(x-5)(x-9) = 0 \implies x = 5 \text{ or } x = 9$.

Next, we must check these potential solutions against the domain of the original logarithmic terms:

• For $x = 5$, the term $\log_5(2x-10)$ becomes $\log_5(0)$, which is undefined because the argument of a logarithm must be strictly positive. Thus, $x = 5$ is an extraneous root.
• For $x = 9$, both arguments $x^2-4x-5 = 81-36-5 = 40 > 0$ and $2x-10 = 18-10 = 8 > 0$ are positive. Thus, $x = 9$ is the only valid solution.