Solve the inequality
$$\frac{\log_ax}{x^2+(a-4)x+4-2a}\le0$$
My work:
If $1>a>0$ then $x \in (0;1] \cup (2; + \infty)$ .
If $3>a>1$ then $x \in (0;2) $ .
If $3 \le a<4$ then $x \in (a-2;2) \cup (0; 1]$ .
If $a=4$ then $x \in (0;1]$ .
If $a>4$ then $x \in (0;1] \cup (2; a-2)$ .
Is it correct?
Firstly from $\log_ax$ you have $a \ne 1$.
Then the inequality is equivalent to: $$\log x \cdot \log a \cdot (x-2) \cdot(x+a-2) \le 0$$ and $$(x-2)(x+a-2) \ne 0$$
Then you could discuss on $a$'s value and write up the solution.