Is $|x^r|=|x|^r$ for real numbers $x$ and $r$?

Question

Suppose $x$ and $r$ are real numbers. Is $|x^r|=|x|^r$? If so, how do you prove it at the lowest level? (That is, using definitions and theorems available at K-12 level. If this is not sufficient then please feel free to rise to the appropriate level.)

Answer 1

Sorry, I don't know what K-12 level means. But evaluating $x^r$ for real $r$ and negative $x$ really only makes sense in rather few cases, as long as you stick to real numbers. Is does make sense if $r$ is a positive integer, and in this case the proof should be obvious.

It does also make sense if $r=p/q$ with $p$ a positive integer and $q$ a positive odd integer. In that case $x^r$ is the $q$-th root of $x$ to the power of $p$ and again, the proof of the equation is easy.
If $r$ is either of the form $p$ or $p/q$, $p$ a negative integer and $q$ a positive odd integer, the proof is also easy.

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Edit: Now for the general case, where we understand complex numbers but only allow real values for $x$ and $r$.

If we do not restrict ourselves to real numbers, then $x^r$ is in general not unique, but it is of the form $e^{r\ln x}$, where there are in general infinitely many choices for $\ln x$. However, if $r$ is real, then $|e^{r\ln x}|$ is actually unique, and is equal to $e^{r\cdot\mathrm{Re}(\ln x)}$. Now, the real part of $\ln x$ is $\ln|x|$. This immediately shows that $|x^r|=|x|^r$.

Answer 2

Let's restrict first to real $x$:

There is a pesky special case to consider first: If $x=0$, then only $r>0$ makes mathematical sense, in which case: $$ |x^r| = |0^r| = |0| = |0|^r = |x|^r.$$

Next assume $x \neq 0$, and rational $r = p/q$ where $p$ is a whole number and $q$ is a positive odd whole number, so that $x^r$ is guaranteed to be a real number. Recall, for all real numbers $x$, it is true that $|x| = \sqrt{x^2}$.

$$ |x^r| = \sqrt{ (x^r)^2 } = \sqrt{ x^{r\cdot 2} } = \sqrt{ x^{2r} } = \sqrt{ (x^2)^r } = (\sqrt{x^2})^r, $$ where the last equality only works because $x^2 > 0$. Then by the same trick as above, $$ (\sqrt{x^2})^r = |x|^r. $$

The same proof works when $x >0$ and $r$ is an arbitrary real number. All that we require is that $x^r$ is real.

Hope this helps!