So I'm just going through some basic algebra as a refresher, and I'm wondering why a number taken to a certain power (such as $2^3$) has an additional negative value if the exponent is expressed as a fraction. For example, $2^{6/2}$ is equivalent to $\sqrt{2^6}$, which equals $\sqrt{64}$, which equals $\pm 8$.
Why the $\pm 8$ for $2^{6/2}$ if $2^3 = 8$?
On
There are two notions of a square root, one as a real function from $[0,\infty) \rightarrow [0,\infty)$ given by $x \rightarrow \sqrt{x}$ and the other as solutions to the equation $x^2=a$. In the first case we exclude negative values in domain and codomain so we don't have to consider them. In the second the solutions are $\pm \sqrt{a}$ and so two square roots. In this case since we have no mention of quadradic equation I would interpret it as the positive square root and conclude that $2^{\frac62} = 2^3=8$.
On
$\sqrt{x}$ is the "principal square root", which is by definition positive, so $\sqrt{64}$ would be $8$. More generally, $\sqrt{x^2} = |x|$.
The plus and minus stuff usually comes in when we're faced with something like $x^2 = 4$, where there are two possible answers to $x$.
So when solving that, we take the square root of both sides: $\sqrt{x^2} = \sqrt{4}$ which is the same as $\sqrt{x^2} = 2$ which is the same as $|x| = 2$, and by definition $|x|$ is $x$ if $x \geq 0$, or $-x$ if $x < 0$, which is why we get two results in situations like that.
Meanwhile $\sqrt{2^3} = |2^3| = |8| = 8$, since $|8| = 8$ due to $8 > 0$ holding, meanwhile $|8| = -8$ doesn't because $8 < 0$ is false, so we only get one value.
Different people will define things in a different way, but I'd say that $ 2 ^ { 6 / 2 } $ is only $ 8 $, since $ 6 / 2 = 3 $ so $ 2 ^ { 6 / 2 } = 2 ^ 3 $. In fact, I'd also say that $ \sqrt { 6 4 } $ is only $ 8 $, since $ \surd $ means the principal square root, not any square root. That second choice is more controversial, and there are times when it's useful to let $ \surd $ mean any square root (although you can also write $ \pm \surd $ to achieve this), and I've seen people define things that way. But the first choice, where $ 2 ^ { 6 / 2 } = 2 ^ 3 $, is necessary for the rule that you can substitute equal expressions wherever they occur, which is fundamental to algebra, and I don't know anybody who would do it differently.
In fact, if you're working from a reference that says that $ b ^ { m / n } $ always means the same as $ \sqrt [ n ] { b ^ m } $, then I'd check the fine print: are they claiming this for any natural numbers $ m $ and $ n $, or only when $ m $ and $ n $ are relatively prime (meaning that the fraction $ m / n $ is in lowest terms)? I'd be surprised if they really mean that it's true even when $ m / n $ can be reduced, at least when $ b $ can be any real number. (Although you could say that $ b ^ { m / n } = \sqrt [ n ] { b ^ m } $ even when $ m / n $ can be reduced, if you also say that $ \sqrt [ n ] { \vphantom 0 } $ means only the principal root like I would, and you require $ b \geq 0 $, so that might be in the fine print instead.)