The domain of $\ln(x)^{\ln(x)}$ I'm a little bit confused!
what is the domain of this function:
$$
\ln(x) ^{ \ln(x) }
$$
this function, in fact, is:
$$
\exp(\ln(\ln(x))\cdot\ln(x))
$$
 so the domain would be:
$$
x>1
$$
But:
$x$ can "also" take on the value (for example) :$$\sqrt[3] {e^{-1}}$$
then there would be some other numbers, not included in $x>1$, in the domain of the function.
 A: Analysts like continuous functions.  Thus for this problem they would either use domain $x>1$ or else use complex numbers, taking a "principal value" for the logarithms.  
Recreational mathematicians like to do things like $\sqrt[3]{e^{-1}}$ for this.  But there is no known use of that type of calculation in mathematics itself.
A: We can avoid the logs for a start and ask about the domain of $y^y$, then set $y=\ln x$ as $\ln y$ is surjective on the reals.  $y^y$ is defined whenever $y \gt 0$ or $y$ is a negative integer, so its domain is $k(k \lt 0, k \in \Bbb Z) \cup k(k \in (0, \infty), k \in \Bbb R)$  That would give the domain of $\ln x^{\ln x}$ as $e^{k}(k \lt 0, k \in \Bbb Z) \cup e^k(k \in (0, \infty), k \in \Bbb R)$ or $e^{k}(k \lt 0, k \in \Bbb Z) \cup m(m \in (1, \infty), m \in \Bbb R)$  
More generally, we write $a^b$ with domain $a \gt 0, b\in \Bbb R \cup a \lt 0, b \in \Bbb Z $ (except $0^0$) but it makes some sense to think of the two pieces of the domain as separate functions.  The first is $e^{b \ln a}$ and has a nice contiguous domain.  The second is $(-1)^be^{b \ln(-a)}$ and because of the $(-1)^b$ piece we can only do it by repeated multiplication, which forces $b$ to be integral.  One obstacle to seeing this is that $a \gt 0, b \in \Bbb Z$ belongs to both pieces of the domain.  Generally in a problem you will only use one piece of the domain or the other. It will be clear from context which you are using and you can pretend that the other piece of the domain does not exist for that problem.
A: If you are more interested in continuity than in algebraic properties then you might want to extend $\ln x$ analytically so that it is equal to $\ln|x|+\pi i$ for negative real numbers.
