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.

Logarithmic equation:
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0 $$
In general:
$$\log_{f(x)}(g(x))\geq 0 $$
The general solution to this has two cases.
Case 1:$$ f(x) \in {(0,1)} \cap g(x) \in {(0,1]}  $$
Case 2:$$ f(x) \in {(1,\infty)} \cap g(x) \in {[1,\infty)}  $$
Solution Set:
$$ \implies x \in {(CASE 1) \cup (CASE2)}  $$
This will give the desired result for the demand of the logarithm to be non-negative. Comment if you want to know how the cases arise.
Final Solution: $$\displaystyle x\in \left[3,4\right)\cup \left(4,5\right)\cup \left(5,\infty\right)$$
http://www.wolframalpha.com/input/?i=Solve+the+inequality:+%5B%2F%2Fmath:log((x%2F5),+(x%5E2-8x%2B16))%3E%3D0%2F%2F%5D
