For even integers $a$ and $b$, since $(x^b)^{1/a}$ and $(x^{1/a})^b$ have different domains, does it follow that $x^{b/a}$ is not well-defined?

Question

Let $f(x)=(x^b)^{1/a}$ and $g(x)=(x^{1/a})^b$, where $a,b$ are even integers.

The domains are clearly different. $\operatorname{dom}\{ f\}=\mathbb{R}$, while $\operatorname{dom}\{ g\}=\mathbb{R}^+_0$.

So, this should mean that $h(x)= x^{b/a}$ is not well-defined, right?

Answer 1

Please note that a rational exponent is defined as follows.

Let $r=\frac{m}{n}$, where $m,n\in \mathbb{Z}$, be a rational and $x$ be any real number. $x^r$ is defined as$$x^{r}=x^{\frac{m}{n}}:= (x^{\frac{1}{n}})^m=(\sqrt[n]{x})^m.$$This definition is always well-defined for all $x$ in the domain of the the function $x^{\frac{1}{n}}$; the domain is $\mathbb{R}$ if $n$ is odd, and the domain is $[0, \infty )$ If $n$ is even.

With this definition, the following property always holds for all $x$ in the domain: $$x^{\frac{m}{n}}=(x^{\frac{1}{n}})^m=(x^m)^{\frac{1}{n}}.$$