Doubt with $\sqrt[3]{x} \ne x^{\frac{1}{3}}$ In a test, there was the following question: What is the value of $(-0.125)^{\frac{1}{3}}$?
One of the possible answers was "$-0.5$" and another answer was "None of the above".
It is important to note that we are only working on real numbers
Most of the students marked the first as the correct answer, but I am pretty sure that $\sqrt[3]{x} \ne x^{\frac{1}{3}}$ because it isn't good defined. For instance:
$(-8)^{\frac{2}{6}}=\sqrt[6]{(-8)^2} = \sqrt[6]{64} = 2$
or
$(-8)^{\frac{2}{6}}=(-8)^{\frac{1}{3}}=\sqrt[3]{(-8)}=-2$
Also, if i graph $f(x)=x^{\frac{1}{3}}$, it plot that, apparently, $(-0.125)^{\frac{1}{3}}=-0.5$.
Wolfram Alpha puts  $\sqrt[3]{x} = x^{\frac{1}{3}}$ for any number that i tried, but another calculator, Photomath, show the problem with the name of "Indeterminated"
Is my approach correct?
Why different calculators or math engines gives different answers for this problem?
 A: If $n$ is odd, $x^n$ is an invertible function from $\Bbb R$ to itself; we denote the inverse either $\sqrt[n]{x}$ or $x^{1/n}$. This is consistent with $\left(x^a\right)^b=x^{ab}$. We can now uniquely define $x^{p/q}\in\Bbb R$ for any $x\in\Bbb R$ with $x\ne0$ (a restriction we can drop if $p/q\gt0$), and any integers $p,\,q$ with odd $q>0$. It won't work for $p$ odd and $q$ even if $x<0$, but in other cases we can cancel $p/q$ into the odd-$q>0$ form. So in general, we define $x^{p/q}$ by cancelling the exponent into its lowest terms first.
A: I think it is $\sqrt[3]{x}=x^{\frac{1}{3}} \ne x^{\frac{2}{6}}$
Consider this example $-1=(-1)^3=(-1)^{2.\frac{3}{2}} \ne ((-1)^2)^{\frac{3}{2}}=1$
The idea is that we must write the numbers we use in their irreducible forms to remove any kind of ambiguity.
A: In fact it is not unproblematic to work with an expression $a^x$ for $x \notin \mathbb Z$
and $a <0$. See my answer to Why $(-2)^{2.5}$ isn't equal to $((-2)^{25})^{1/10}$?
The "universal definition" would be $a^x = e^{x\ln a}$, but is valid only for $a > 0$. Extending this to $a < 0$ is possible, but involves the complex logarithm and doing so the problem is that it has infinitely many branches. Note that $w = \ln z$ must satisfy $e^w = z$, but this equatuon has infinitely many solutions in $\mathbb C$. If you pick any solution $w_0$, then the complete set of solutions is given by the $w_k = w_0 + 2\pi i k$, $k \in \mathbb Z$.
Usually one takes the principal branch on $\ln z$ which is characterized by the condition that $0 \le \arg(\ln z) < 2 \pi$. This has the benefit that $\ln z$ is the standard real logarithm if $z$ is a positive real number. With that approach we get $\ln (-1) = \pi i$ (because $e^{\pi i} = -1$) and $(-1)^{1/3} = e^{\pi i/3} = 1/2 + i \sqrt{3}/2$. This differs from $\sqrt[3]{-1} = -1$. But note that $1/2 + i \sqrt{3}/2$ is one of three complex cubic roots of $-1$.
If $x = p/q$ with $p \in \mathbb Z$ and $q \in  \mathbb N$, then I nevertheless think it is acceptable to define for $a < 0$
$$a^{p/q} = \sqrt[q]{a^p}$$
provided $q$ is odd. See again Why $(-2)^{2.5}$ isn't equal to $((-2)^{25})^{1/10}$? This can be viewed as making a special choice for $\ln a$ depending on $x$. In fact, we may take $\ln a$ such that $\arg(\ln a) = q\pi$.
