Is $|x^r|=|x|^r$ for real numbers $x$ and $r$? 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.)
 A: 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.

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 $e^{r\cdot\mbox{Re}(\ln x)}$.
Now, the real part of $\ln x$ is $\ln|x|$.  This immediately shows that $|x^r|=|x|^r$.
A: 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!
