Several values of irrational exponentiation When talking about a number to a rational exponent, there are as many answers as the denominator of the exponent. Like the question: Is $9^{1/2}$ equal to $3$ or $-3$.
However when we have an irrational exponent like $2^{\sqrt2}$, I cant immediatly see more than one candidate solution: $e^{\sqrt2\log2}$. Are there any ways to interpret irrational exponents such that they give several candidate solutions?
Another way to state this question would be:

are there more than one complex solution to $x^{1/p}=y$ for irrational $p$ and positive real $y$.

 A: In the real numbers field we have no ambiguity if we define (as usual) $a^{\frac{1}{2}}$ as the positive root of $a$ and $a^\alpha=e^{\alpha \log a}$ (for $a\le 0$) .
The problems come out when we go to complex numbers, as a consequence of the fact that $e^{i 2 k \pi}=1 \;\forall k \in \mathbb {Z}$. So $y=e^{i x}$ is a periodic function ( also for $x \in \mathbb{R}$) and is not invertible.
So, in the complex numbers field we have:
$$
y=e^x \iff y=e^x\cdot 1 = e^xe^{i2k\pi}=e^{x+i2k\pi}
$$
and the inverse '' function'' is a multivalued function : 
$$
\log y= x+i2k\pi
$$
we can find $x$ writing the complex number $y$ in polar form as $y=|y|e^{i\theta}$ so, from $|y|e^{i\theta}=e^x$ we have $x=\log |y| +i\theta$ and:
$$
\log y= \log |y| +i\theta+i2k\pi
$$ 
As a consequences of this fact  for $y,x,\alpha \in \mathbb{R}$ and  $y,x >0$ (this imples $\theta=0$), we have:
$$
y=x^\alpha \iff y=e^{\alpha\log x}=e^{\alpha(\log |x|+i2k\pi)}
$$
For $\alpha=1/2$ this gives the two square roots and for $\alpha$ irrational gives, in general, many values.
