$(a^m)^n$ with negative $a$

Question

I have a question about the following exercise:

Evaluate $$\left(\left(-2\right)^2\right)^{\frac{1}{2}}.$$

When I solve it I get $2$. I first raise $-2$ to the second power, which is $4$. After that I raise four to the $1/2$ power, which is the same as taking the square root of it. So I get $2$.

But when I put this question in Symbolab, I get $-2$. They are using a formula of exponents. I think their answer is wrong, especially because many other sites, for instance Wolfram Alpha, are saying the answer is $2$.

Link Symbolab exercise: https://www.symbolab.com/solver/step-by-step/%5Cleft(%5Cleft(-2%5Cright)%5E%7B2%7D%5Cright)%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D

Is it correct that when you want to use the formula underneath, $a$ has to be a positive number? Or is there another reason why this rule doesn't seem to work in my example? $$(a^m)^n = {a}^{mn}$$.

Welcome to Mathematics Stack Exchange. Fractional powers of negative numbers are not uniquely defined, and the "general rule" $(a^m)^n=a^{m×n}$ does not always work when $m$ and $n$ are not integers. Cf. this question — J. W. Tanner, Feb 24 '20 at 20:52

score 1 · Accepted Answer · answered Feb 24 '20 at 20:51

Yes, to use the formula $(a^b)^c = a^{ac}$ completely without issues, you need to meet at least one of the following conditions:

The exponents are integers
The base is a positive number

This is because negative bases raised to fractional exponents are, if not impossible, at least cumbersome and fraught with pitfalls. So they are, in my opinion, better to simply avoid completely.

$(a^m)^n$ with negative $a$

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