Properties of Logarithms

We can rewrite $\log_4{(x-2)}$ as

$\log_4{(x-2)} = \frac{\log_2{(x-2)}}{\log_2{4}} = \frac{1}{2} \log_2{(x-2)}.$

Since $1 = \log_2{2}$ , the given equation can be rewritten as

$\log_2{(x-5)} = \frac{1}{2} \log_2{(x-2)} + \log_2{2}.$

Multiplying 2 on both sides, rearranging terms, and factorizing, we have

$\begin{aligned} \log_2{(x-5)} &= \frac{1}{2} \log_2{(x-2)} + \log_2{2} \\ 2\log_2{(x-5)} &= \log_2{(x-2)} + \log_2{4} \\ \log_2{(x-5)^2} &= \log_2{4(x-2)} \\ (x-5)^2 &= 4(x-2) \\ 0 &= x^2-14x+33 \\ 0 &= (x-3)(x-11). \\ \end{aligned}$

Thus, $x=3$ or $x=11$ . However, $x=3$ cannot be a solution because $x-5 = -2 < 0$ , and the domain of the logarithm function is positive numbers.

Therefore, $x = 11.$