Solve logarithmic equation $\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$ Find $x$ from logarithmic equation:
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0 $$ 
This is how I tried:
$$x^2-8x+16>0$$
$$ (x-4)^2>0 \implies x \not = 4$$
then
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq \log_{\frac{x}{5}}(\frac{x}{5})^0 $$ 
because of base $\frac{x}{5}$, we assume $x \not\in (-5,5)$, then
$$x^2-8x+16 \geq 1$$
$$ (x-3)(x-5) \geq 0 \implies$$
$$ \implies x \in {(- \infty,-5) \cup (5, \infty)} \cap x\not = 4 $$
But this is wrong, because the right solution is $$x \in {(3,4) \cup (4,6)} $$
I'm sorry if I used the wrong terms, English is not my native language.

Given $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\;,$$ Here function is defined when $\displaystyle \frac{x}{5}>0$ and $\displaystyle \frac{x}{5}\neq 1$ 
and $(x-4)^2>0$. So we get $x>0$ and $x\neq 5$ and $x\neq 4$
If $$\displaystyle \; \bullet\;  \frac{x}{5}>1\Rightarrow x>5\;,$$ Then $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\Rightarrow (x^2-8x+16)\geq 1$$
So we get $$\displaystyle x^2-8x+15\geq 0\Rightarrow (x-3)(x-5)\geq 0$$
So we get $x>5$
If $$\displaystyle \; \bullet 0<\frac{x}{5}<1\Rightarrow 0<x<5\;,$$ Then $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\Rightarrow (x^2-8x+16)\leq 1$$ 
So we get $$\displaystyle (x-3)(x-5)\leq 0$$
So $$3\leq x<5-\left\{4\right\}$$
So our  final Solution is $$\displaystyle x\in \left[3,4\right)\cup \left(4,5\right)\cup \left(5,\infty\right)$$
