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.

$$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$$
$$=\frac{\log{(x^2-8x+16)}}{\log\frac{x}{5}}\geq0$$
So $x$ is non-positive. And now raising both sides on the base $10$.
$$x^2-8x+16\geq1$$
$$x^2-8x+15\geq0$$
$$(x-5)(x-3)\geq0$$
$$x\geq5; \ x\geq3$$
But $x$ cannot be $5$. So answer is $[3,5)\cup(5,\infty)$
