Find the domain of the following function $ f(x)={(\ln x)}^{\ln^{2}x} $. I know that $ x \gt 0 $ because of logarithm precondition, and I can see that $ x \neq 1 $ because otherwise it would lead to $ 0^0$ which is problematic, but when I checked the graph of the function I have discovered that it started from $ 1 $ on $ x $ axis and $ (0,1) $ interval is not considered.
In sum, I thought that the domain should have been $ (0, \infty) - \{1\}$ but it seems to be $(1,\infty)$ and I can't figure out where I am wrong.
 A: Note: this question has only been addressed at the high school level.  Therefore, the given answer does not address the question from a mathematician's point of view.

What I want to do in this answer is to find the largest interval, such that for all values ​​in this interval the function $f(x)=(\ln x)^{\ln^2 x}$ is well-defined/meaningful.
By definition we have $0<x\neq 1$.
$$\operatorname{dom}\left[(\ln x)^{\ln^2 x}\right]\longmapsto \ln x>0\vee \left\{\ln x<0\wedge \ln^2x\in\mathbb Z\right\}$$
and
$$\ln x>0\implies x>1$$
Then note that, if $\ln x<0$ then, $\ln ^2x \in\mathbb Z$.
Thus,
$$\ln x=-\sqrt n\implies x=e^{-\sqrt n},\, n\in\mathbb Z_{>0}.$$
With all this in mind, we can conclude that,
$$
\begin{aligned}&\operatorname{dom}\left[(\ln x)^{\ln^2 x}\right]\\
=&\left\{x\mid x>1\vee x=e^{-\sqrt n}, \,n\in\mathbb Z_{>0}\right\}
\end{aligned}
$$
If you assume $0^0=1$, then $x=1$ is included in the domain:
$$
\begin{aligned}&\operatorname{dom}\left[(\ln x)^{\ln^2 x}\right]\\
=&\left\{x\mid x\ge1\vee x=e^{-\sqrt n}, \,n\in\mathbb Z_{>0}\right\}\end{aligned}
$$
A: First of all, we can see that $f(e^{-1}) = {(\ln e^{-1})}^{\ln^{2}e^{-1}} = -1$ so we can see that it is indeed defined for some values of $x < 1$ but $f(e^{-1/2})$ is undefined for example.
If you think about it, this question is very similar to asking for the domain of $x^x$ so this might help you. Furthermore, as @David said on his comment, if $a < 0$ then $a^b$ is "problematic". Knowing this, to avoid any problems we would want $\ln x > 0 \implies x > 1$ so we could just conveniently define $\text{dom} f(x) = (1,+\infty)$. And in this case:
$$f(x) = {(\ln x)}^{\ln^{2}x} = (e^{\ln(\ln x)})^{\ln^2 x} = e^{\ln(\ln x) \cdot \ln^2 x}$$
Now let's try to find a larger domain that includes some $x < 1$, to avoid raising numbers to an irrational exponent that is another problem on its own, let's consider all rational $\ln x < 0$, then for some $p,q \in \mathbb{N}$, $0 <x = e^{p/q} < 1$ and thus $$f(x) = f(e^{p/q}) = (p/q)^{(p/q)^2} = \sqrt[q^2]{(p/q)^{p^2}}$$ remember that $p/q < 0$ so for this expression to be properly defined we want $p$ to be even or $q$ to be odd.
So we can as well define the domain of $f(x)$ as $(1,+\infty) \cup \{\ e^{p/q} \in (0,1) \mid p = 2k \text{ or } q = 2k + 1 \text{ for some } k\in\mathbb{Z} \}$.
So its domain really depends on what you define it to be.
