When is it possible to "move" an exponent out of a radical?

Question

It seems when a value in a radical is positive it's valid to "move" the exponent out of the radical.

Consider the function $\sqrt{x^2}$. When $x\geq0$ then

$\sqrt{x^2} = (\sqrt{x})^2$

For example, when $x = 5$

$\sqrt{5^2} = (\sqrt{5})^2$

$\sqrt{25} = \sqrt{5} \cdot \sqrt{5}$

$5 = 5$

However, when $x<0$ it's no longer valid to "move" the exponent out.

For example, when $x=-5$

$\sqrt{(-5)^2} \not= (\sqrt{-5})^2$

$\sqrt{25} \not= \sqrt{-5} \cdot \sqrt{-5}$

$5 \not= -5$

Also, it seems if a value inside a radical can be rendered positive, then an exponent can be "moved" out.

For example,

$\sqrt{(x^2)^3} = (\sqrt{x^2})^3 = |x|^3$

Thus, is it correct to say that if a value inside a radical is positive or can be rendered positive, an exponent can be "moved" out from a radical?

I'm guessing if this is the case, then it has something to do with the product rule of radicals which requires the values inside the radicals to be positive beofre being "combined":

if $a \ge 0$ and $b \ge 0$, then $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$

Answer 1

Yes, it is correct, and it follows from the properties of powers:

$$(x^{\alpha})^{\beta}=(x^{\beta})^{\alpha}$$

whenever all the quantities involved are well defined, e.g. for positive $x$ and real $\alpha,\beta$.

Answer 2

$x^r$ is well defined for all $x\ge 0$ and $r\in\mathbb R$ (except maybe the case $x=0$ and $r\le 0$).

In particular if $r=\frac pq$ is rational then

$$x^r = x^\frac pq=\sqrt[q]{x^p}=(\sqrt[q]{x})^p$$

$x^r$ is also well defined for $x<0$ and $r\in\mathbb Q_\text{odd} \{ (p,q)\in\mathbb Z^2\mid \gcd(p,q)=1\text{ and } q \text{ odd}\}$.

In this case since $x\mapsto x^q$ is an odd function, then its reciprocal is defined everywhere by $\sqrt[q]{x}=\operatorname{sign}(x)\sqrt[q]{|x|}$

In this case also we can swap the order of $p$ and $q$ in the calculation.