Q: How exactly is the function $x^a$ defined? I need some refresher here since the more I think about this the more I get confused. 
How exactly is the function $f(x) = x^a$ defined (a - real constant, $x$ - real variable)? 
1) 
What is its domain? E.g. if $a=1/3$ then I know we can plug any values of $x$ so it seems the domain is the full set of all reals in this case. Is this correct? 
But if we have e.g. $a=1/4$ then only non-negative values for $x$ are allowed.
Also... to prove that the function $x^a$ is continuous in real analysis they often use the representation 
$$x^a = b^{a \cdot log_b{x}}$$ for some $b > 0, b \neq 0$
and then refer to the fact that the functions $b^x$ and $log_b{x}$ and the constant function $a$ are all continuous. But if so... this proof of continuity is incomplete, because for negative values of $x$ we cannot use this representation.
Or maybe... is it the case that even when $a=1/3$ (in real analysis) we still assume $x$ has to be positive? I think my real analysis (university) book assumes just that but then... I remember from high-school that I can take cubic roots of negatives so... Where is the catch?  
2) 
In relation to 1)... how are we supposed to understand this expression/function $x^{\sqrt3}$? Does that expression require $x$ to be positive in order to have any meaning? 
I mean, for negative $x$ there is a similar problem here. We cannot just approximate $\sqrt3$ with ANY converging sequence of rationals $\frac{p_k}{q_k}$ (for k=1,2,3,...) and then define that $x^{\sqrt3}$ is the limit of the corresponding sequence:
$$x^\frac{p_1}{q_1}, x^\frac{p_2}{q_2}, ..., x^\frac{p_n}{q_n}, ...$$
Why? Because some of these rationals $\frac{p_i}{q_i}$ may have even denominators and then that means $x^\frac{p_i}{q_i}$ is not defined (when x is negative).
3) 
All these thoughts bring me to the ultimate confusion. 
What is really the co-domain of $g(x) = x^{\frac{2}{6}}$ (or say what is its value for $x=-27$)? 
Is this function equal to $\sqrt[6]{x^2}$ and thus generating only positive values no matter what sign $x$ has... or is it $x^{\frac{1}{3}}$ i.e. $\sqrt[3]{x}$ and thus generating both positive and negative values?  

A: Common convention is


*

*no problem with positive basis (and the log representation is fine);

*no problem with zero basis and positive exponent;

*$0^0$ can be $0$ or $1$ depending on contexts; negative power not allowed;

*if the basis is negative,


*

*a rational exponent must be written in simplified form and have an odd denominator. The rule $x^{m/n}=\sqrt[n]{x^m}=(\sqrt[n]x)^m$ works.

*an irrational exponent is not allowed.

If you allow complex answers, then


*

*rational powers define $n$ distinct branches,

*irrational powers define a principal branch $\sqrt[m/n]{-x}\text{ cis}(\frac{m\pi}n)$ or are not allowed.

A: While it is somewhat common to interpret the possibility of dealing with a negative base and non-integer exponent, there are several issues. The entire situation leads to a loss or restriction of many nice exponential properties, as well as things such as continuity such as what you've mentioned. This means things like $x^{ab}=(x^a)^b$ needn't be true, which is why $x^{2/6}$ doesn't make much sense to you. And in the case of something like $x^{\sqrt3}$, you can see an even worse issue. If we let $x^{\sqrt3}=\lim\limits_{r\to\sqrt3}x^r$ over rational $r$ with odd denominators, the resulting limit still doesn't exist because the result can be positive or negative depending on the parity of the numerator.
Often when jumping into roots with negative bases, we necessarily jump into the complex plane, which allows us to keep most of the properties desired and avoid passing off half the cases because of negative bases. There we still define $x^a$ as $\exp(a\log(x))$, but we either use what's called a principal branch, or we consider a multivalued logarithm. The latter allows us to interpret $1^{1/2}$ as $\pm1$ for example, while the former gives us a well-defined exponentiation that's continuous everywhere except for on the branch cut. If one chooses to go with the well-defined approach, then one will end up with non-real results, such as $(-1)^{2/3}\simeq-0.5 +0.866i$ instead of $(-1)^{2/3}=1$ as you might expect.
But if we're going to be sticking to the reals, then the question is what use does involving negative bases give us? It only complicates the results, and if we're going to have $x^{a/b}$ to be well-defined when $x<0$ and $b$ is odd, then the resulting function is either an even or odd function, so it's just a reflection of the function for $x>0$.
Best to define $x^a$ as a function from the positive reals to the positive reals and simply avoid all of the above complications.
