# Integer solutions to the inequality $\log_{1/5}\log_3\frac {x-3}{x+3}\ge0$

$$\log_{1/5}\log_3\frac {x-3}{x+3}\ge0$$

If $x$ is of the interval $[-8,10]$

Now I solved this, tried to limit $x$ as much as I could but I consistently get that there should be $10$ values of $x$ for which this holds. The answer is a measly $3$, so I'm missing an important step.

I limited $x$ to $(-\infty, -6]\cup(-3,+\infty)$, whatever help that is to any of you.

The function $\log_{1/5}$ is decreasing, so your inequality is equivalent to $$0<\log_3\frac{x-3}{x+3}\le 1$$ which in turn is equivalent to $$1<\frac{x-3}{x+3}\le 3$$ The left hand side inequality is $$\frac{x-3}{x+3}-1>0$$ or $$\frac{-6}{x+3}>0$$ that gives $x<-3$.
The right hand side inequality is $$\frac{x-3}{x+3}-3\le0$$ or $$-2\frac{x+6}{x+3}\le0$$ which gives $x\le-6$ or $x\ge-3$.
Putting together the two inequalities gives, as solutions, $x\le-6$.
Note that your interval $x>-3$ is wrong. For instance, if $x=0$ you get $$\frac{x-3}{x+3}=-1$$ and the logarithm is not defined. You forgot the condition that $\log_{1/5}t$ is defined only for $t>0$.