confused about law of exponent (contradiction) Is the following statement true?
$$
(-1.2)^{1.6} = \sqrt[10]{(-1.2)^{16}}.
$$  
Is it because $a^{0.5} = a^{1/2} = \sqrt{a}$?

I'll take $\sqrt{}$ as example because for odd numbers, you get a bijection from $\mathbb{R}$ into itself (with positive number mapped to positive numbers and negative numbers mapped to negative numbers) so the problem only arises with even numbers and once you understand for $\sqrt{}$, you'll understand for $\sqrt[4]{}$, $\sqrt[6]{}$ etc. as well.
You problem is that you didn't understand the function $\sqrt{} : \mathbb{R_+} \to  \mathbb{R_+}$ is just a convention.
When we write $\sqrt{4}$, we mean $2$ because we chose the positive real number as result for the function $\sqrt{}$ because if we didn't specify that, we wouldn't get a function since a function is something that has a unique value for each "input".
But if you try to extend that function to $\mathbb{R}$ we get $\sqrt{} : \mathbb{R} \to ?$ where does it go? Because now, for $\sqrt{-1}$, you have to decide between $i$ and $-i$...
You would pick $i$ because it looks positive but it is not... There has been a question about a way to differentiate $i$ from $-i$ recently on SO and the answers were, more or less, that you can't.
Of course, for $\mathbb{R}$, you could decide to always take the one that is a positive real number multiplied by $i$. But then you get $\sqrt{} : \mathbb{R} \to \mathbb{C}$ and you only use $i$ in $\mathbb{C}$ so you don't get a bijection.
And then if you try to extend it to $\mathbb{C}$ with $\sqrt{} : \mathbb{C} \to ?$ you can't decide which one to take unless you define some kind of sign over $\mathbb{C}$ to partition it. Which you usually don't want to do.
Plus when you work in $\mathbb{C}$, you usually just want one of the answers, of $a^2=b$ and you don't care which one you get.
