Why can't $x$ be negative in $x^{\ln{y}}$

Question

According to wolfram alpha, the domain of $x^{\ln{y}}$ is $x>0$ and $y>0$ but putting $(-1,1)$ for $(x,y)$ I get a perfectly fine answer of 1?

$y>0$ makes sense, since $\ln{y}$ is only defined there. But why must $x>0$ as well. The only other problem I see is $x = 0, y = 1$ as then we get a $0^0$ which is not defined.

I can't see why $x>0$! (exclamation, not factorial)

Answer 1

When $x < 0$, the result of $x^{\ln y}$ is only real at some cherry-picked points. I suppose WA wants to make it a real function.

Answer 2

Note that:

$x^{ln(y)}=(e^{ln(x)})^{ln(y)}=e^{ln(x)ln(y)}=(e^{ln(y)})^{ln(x)}=y^{ln(x)}$

Therefore, if you accept that $y > 0$ you should also accept that $x > 0$ as well