Does the identity $|x^n|^{\frac{1}{n}} = |x|$ hold?

Question

Does $|x^n|^{\frac{1}{n}} = |x|$? Also n is a natural number. Sorry that this is such a stupid question, I'm just simplifying something and trying to make sure I'm doing it right.

Answer 1

You are correct! you can also show the result by using the definition of $|x|$ and splitting the cases.

Answer 2

Yes (provided $n\neq0$), since $|x^n|=|x|^n$, and $(r^a)^b=\exp(ab\ln r)=r^{ab}$ holds for all real $r>0$ and all $a,b$, while $(0^a)^b=0^{ab}$ also holds in the here relevant case that $a=n$ and $b=\frac1n$.

(Indeed $(0^a)^b=0^{ab}$ holds in all cases where both sides are defined, i.e., for real $a,b\geq0$: both sides are $0$ unless $ab=0$ in which case both sides are $1$. But I'm sure someone is going to attack me for that last case.)

Answer 3

$$\quad \left|x^n\right|=\left|x\right|^n\quad \mathrm{so} \quad\sqrt[n]{\left|x^n\right|}=\sqrt[n]{\left|x\right|^n}=|x|$$