Where am I going wrong in in solving this inequality?

Question

I have to find the domain of

$$\sqrt{\log_{0.5}\frac{x}{x^2-1}}$$

As the under root quantity must be positive,

$$\frac{x}{x^2-1}\leq1\implies\frac{x^2-x-1}{(x+1)(x-1)}\geq0\\\implies x\in(-\infty,-1]\cup\left[\frac{1-\sqrt5}{2},1\right]\cup\left[\frac{1+\sqrt5}{2},\infty\right).$$

But this in incorrect. Where is it wrong?

Answer 1

$\log(0.5) < 0$, so logs to base 0.5 have reversed signs.

Answer 2

You have to solve two inequalities: $$0<\frac{x}{x^2-1}\leq1.$$ The first one is necessary because the argument of the logarithm should be positive. So according to your partial solution, you should remove the set $(-\infty,-1]\cup [0,1]$ and therefore the domain is $$\left[\frac{1-\sqrt5}{2},0\right)\cup\left[\frac{1+\sqrt5}{2},+\infty\right).$$

Answer 3

Alternatively: $$\log_{0.5}\frac{x}{x^2-1}=\log_{2} \frac{x^2-1}{x}\ge 0 \overbrace{\Rightarrow}^{x\ne 0,\ \ x\ne \pm1} \\ \frac{x^2-1}{x}\ge 1 \Rightarrow \frac{x^2-x-1}{x}\ge 0 \Rightarrow \\ \frac{1-\sqrt5}{2}\le x<0 \ \ \text{or} \ \ \frac{1+\sqrt5}{2}\le x. $$