What's the domain of $(\log_{\frac{1}{2}}{x})^x$?

Question

What's the domain of $$(\log_{\frac{1}{2}}{x})^x$$ I know that for defining $\log_{a}{x}$, $x$ must be grater than zero ($x\gt 0$) and ($a\gt$ and $a\neq 1$) but I asked this question in Quora and everyone had different answers with no consensus. Can you solve and explain it?

Answer 1

Here are the constraints:

1. the logarithm is well-defined for $x > 0$.
2. Every function of the form $f(x)^{g(x)}$ is well defined for:
   i. a non-negative base if $g(x)$ is not an integer;
   ii. any real base when $g(x)$ is an integer.

Requirement 2.i boils down to solving $\log_{\frac12} x \geq 0$: $$\log_{\frac12} x \geq 0 \implies -\frac{\ln x}{\ln2} \geq 0 \implies \ln x \leq 0 \implies x \leq 1.$$

The condition 2.ii allows us to pick any $x \in \mathbb Z$.

Putting these two conditions together with the first one, we have the system $$\begin{cases} x > 0\\ x \leq 1 \lor x \in \mathbb Z \end{cases}$$ which, solved, yields $\operatorname{dom} f = (0, 1] \cup \mathbb Z^+$.

Answer 2

Hint:

Understand and prove the following:

$$x>0\;\;\text{and also}\;\;\log_{\frac12}x>0\implies x<1$$

Answer 3

To define the domain you must start from the more '' external'' function, that is the exponential, requiring: $$ \log_{\frac{1}{2}}(x)\ge 0 $$ that you can write as: $$ \dfrac{\log (x)}{\log(\frac{1}{2})}\ge 0 $$ since $\log(\frac{1}{2})<0$ and $\log (x)$ is a real number only if $x>0$, this give: $0<x\le 1$ that is the domain.