Why wolfram alpha assumed $ x>0$ as a domain of definition for $x^x $?

Question

I'm confused why wolfram alpha defined the function $x^x $ for $x>0$ , only what i think that it were defined it as :$e^{x\log x}$ , but the domain of definition depend to the function form .then : Why wolfram alpha assumed $ x>0$ as a domain of definition for $x^x $ ?

Note: for example $x$ and $e^{\log x}$ are equivalent form but they don't have the same domain of definitin .

Edit: I have edited the question just to give an example without changing the meaning of the question

Answer 1

$$e^{x\log x}$$

is defined where the exponent $x\log x$ is defined. Even if $x$ is defined on the whole real line, the function $\log x$ is defined only on the open set $]0,\infty[$. Hence the natural domain is such interval, i.e. $x>0$.

EDIT (The natural domain of $x^x$ without using the form $e^{x\log x}$).

The function $b^x$ is defined if and only if $b>0$. Now you may consider the function $b(x)^x$, where $b$ depends on $x$, and also in this case $b(x)$ must be a positive function (this include that can be constant and positive). In particular $b(x)$ may be the identity function, that is $b(x)=x$. Hence $x=b(x)>0$. Hence the natural domain for $x^x$ is $]0,\infty[$.

Answer 2

irrational powers of negative real numbers are not defined, moreover, the fractional powers of negative numbers are not single valued. I think I can write a much clearer answer, but for now, here is a simple demonstration of the problem:

Let us try to evaluate $\left(-\frac m n\right)^{-\frac mn}$ $$\left(-\frac m n\right)^{-\frac mn}=\sqrt[n]{\left(-\frac n m\right)^m}.$$ Now, for an odd $m$, if $n$ is even, the answer is complex, if it is odd, there is a negative real answer.

Now imagine you have a sequence of rational numbers $\{x_n\}$ converging to some negative irrational number $x$ in a way that the parity of the denominator changes through out the sequence. Obviously $x_n^{x_n}$ does not converge, as it oscillates between real and complex numbers. So, how can we define $x^x$?

It doesn't even quit make sense to define this for all the rational numbers, because how would you determine if $(-0.6)^{-0.6}$ is positive or negative. Do we simplify $6/10$ to $3/5$ first which gives us a negative number, or do we not, which gives a positive number?