What would be the domain of the function $\sqrt{\frac{\log_{0.2}(x-1)}{x^2-2x-8}}$? I need to find the domain of the function 
$$f(x)= \sqrt{\frac{\log_{0.2}(x-1)}{x^2-2x-8}}$$
Here's how far I could get:
So clearly for the square root to be real, whatever's inside has to be greater than or equal to 0.
$$\implies \frac{\log_{0.2}(x-1)}{x^2-2x-8} \geq 0$$
$$\implies \frac{\log_{0.2}(x-1)}{(x+2)(x-4)}\geq 0$$
Also, for the logarithm to exist, clearly $x>1$
But how do I proceed? Clearly for this to be greater than 0, both the numerator and the denominator must be of the same sign, i.e, either both have to be positive or both have to be negative. But how do I set those conditions?
 A: Just going by the properties of the logarithm and polynomials.
The denominator is negative for $x \in (-2,4)$ and positive elsewhere (ignoring critical points).
The numerator can be analyzed by noting that $\log_b(x) = \log(x)/\log(b)$. In our case $\log_{0.2}(x-1) = \log(x-1)/\log(0.2)$. Recall that $\log(x \in (0,1)) < 0$. Thus the numerator is defined for $x > 1$, positive for $x \in (1,2)$, and non-positive for $x \geq 2$.
Intersecting these two domains implies that we must have $x \in [2,4)$!
A: Edited answer:
$x=2$ is the critical point of the numerator, i.e., the point where the function changes it's value, whole those of the denominator are $x=-2,x=4$. Employing the wavy curve method for the denominator, we get that it is negative when $x\in (-2,4)$ and positive for $x\in (-\infty,-2)\cup(4,\infty)$. Also numerator is positive when $x\in (0,2)$, zero at $x=2$ and positive when $x\in (2,\infty)$. Now, it is easy to get that the domain is $x\in [2,4)$, using that $x>1$ for the logarithm to exist.
Hope that helps.
