# Solve logarithmic inequality $\frac{2x-x^2}{\log_3(x-5)}>0$

I'm trying to solve the following logarithmic inequality:

$$\frac{2x-x^2}{\log_3(x-5)}>0 \rightarrow\frac{x(x-2)}{\log_3(x-5)}>0$$

I totally have no clue about how to work with this inequality because there isn't a logarithm in the numerator.

I tried doing the following but I don't this that will lead me to anywhere:

$$\frac{3^{x(x-2)}}{x-5}>0$$

As suggested in the comments, I separately check if the numerators and denominators are positive:

$$x-5>0\rightarrow x>5$$

$$3^{x(x-2)}>0\rightarrow x(x-2) > ?$$

Any hints?

• Hint: Separately check where numerator and denominator are positive/negative. Combine results. – gammatester Oct 10 '16 at 14:53

If the log is well defined then $x>5$. But $2x-x^2<x^2-x^2=0$ for $x>5$. Hence the original inequality is equivalent to $\mathrm{log}_3(x-5)<0$, i.e., $x \in (5,6)$.