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

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 $t>0$.

  • $\begingroup$ Right, I didn't think to include the left side of the inequality, and focused only on the right one. Thank you. $\endgroup$ – John Doe May 30 '15 at 19:31

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