Find the integers which do not satisfy the inequality $\log_{1/2}(x+5)^2>\log_{1/2}(3x-1)^2$.
When I did this expanding the squares I got:
$x^2-2x-3>0\implies (x-3)(x+1)>0\implies x\epsilon(-\infty,-1)\cup(3,\infty)$
Also from domain $x\neq-5,\frac{1}{3}$. Hence answer is $6$.
But when I did use the property of $\log$ as follows:
$\log_{1/2}(x+5)^2>\log_{1/2}(3x-1)^2\implies \log_{1/2}(x+5)>\log_{1/2}(3x-1)\implies x+5<3x-1\implies x>3$
What concepts I'm lacking please help.
$$(x+5)^2\geq(3x-1)^2$$ or $$(x-3)(x+1)\leq0,$$ which gives $-1\leq x\leq3$ and we get the answer $\{-1,0,1,2,3,-5\}$.
I solved it in integer numbers, which you wanted.
You can't cancel squares by your way because $$\log_ax^2=2\log_{a}|x|$$
By this way we need to solve $$|x+5|\geq|3x-1|,$$ which gives the same answer.
Another way to explain.
With domain $x\neq-5$ and $x\neq-\frac{1}{3}$ we obtain: $$\log_{\frac{1}{2}}(x+5)^2>\log_{\frac{1}{2}}(3x-1)^2\Leftrightarrow$$ $$\Leftrightarrow2|x+5|<2|3x+1|.$$
Since we not need satisfy, we obtain: $$|x+5|\geq|3x-1|$$ and with $x=-5$ we get the answer.