Let $x>0$ and $c>0$. Does $x^{1/c}$ always exist? We know that for $x >0 $ in real, $x^{1/n}$ is well defined for $n \in \mathbb{N}$, and is called the $n$-th root of $x$. Just wonder if we can replace $n$ with any real positive real number $c$ and still ensure that $x^{1/c}$ exists? 
For any number $x^c$, I have been happily taking $(x^{c})^{\frac{1}{c}}$ to retrieve my original $x$. Just want to confirm if this operation is valid.
EDIT: I was thinking of letting $c = \pi$ or the Euler's constant...
 A: The answer is yes - $x^y$ is continuous in both $x > 0$ and $y > 0$. If $y \leq 0$ things are still good, but if $x \leq 0$ then it gets ugly.
As for a sketch of a proof, if you are ok with the fact that the inverse functions $\exp(x)$ and $\ln(x)$ are continuous for positive values of $x$, then it follows pretty simply:
$x^\frac{1}{c} = \exp(\ln(x^\frac{1}{c})) = \exp(\frac{1}{c}\ln(x))$
So you're taking a continuous function of $x$, multiplying it by a constant, then taking a continuous function of the result. And, nicely enough, you also get:
$(x^\frac{1}{c})^c = (\exp(\frac{1}{c}\ln(x)))^c = \exp(c \times \frac{1}{c} \ln(x)) = \exp(\ln(x)) = x$
Where to get that result you just have to accept that taking the exponential function to a power is the same as multiplying the argument of the exponential function by that value, which follows from the definition of the function.
A: If $x$ is positive and $c$ is real then the expression $x^{c}$ is well defined. When $c$ is rational the definition is based on notion of $n^{\text{th}}$ roots and the properties of exponents hold well. When the value $c$ is irrational then it is not possible to handle the symbol $x^{c}$ by algebraic methods. One of the approaches to define $x^{c}$ is via the use of exponential and logarithmic functions. Thus $$x^{c} = \exp(c\log x)\tag{1}$$ and here you can see the need for constraint $x > 0$ because $\log x$ is defined only for positive $x$.
The same way $x^{1/c}$ is also defined as long as $x > 0$ and $c \neq 0$ (otherwise $1/c$ does not make sense). Just write $1/c = b$ and then you have $x^{1/c} = x^{b}$ and this $x^{b}$ is defined by the equation $(1)$ above. Also the identity $$(x^{c})^{1/c} = x\tag{2}$$ holds for all positive $x$ and all real non-zero $c$. More generally we have $$(x^{a})^{b} = x^{ab}\tag{3}$$ for all $x > 0$ and all real $a, b$. You can prove $(3)$ using definition $(1)$.
