What's the domain of $(\log_{\frac{1}{2}}{x})^x$? 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?
 A: Here are the constraints:


*

*the logarithm is well-defined for $x > 0$.

*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 $\operatorname{dom} f = (0, 1] \cup \mathbb Z^+$.
A: Hint: 
Understand and prove the following:
$$x>0\;\;\text{and also}\;\;\log_{\frac12}x>0\implies x<1$$
A: 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.
