which order do we perform the exponents $x^{2*1/2}$? For $x^{2*1/2}$, do we do $\sqrt x^{2} $ or $ \sqrt{x^{2}}$ ? the reason is because the second one is defined for the entire real line and is equal to |x| whereas the first one is only defined for positive numbers. But at the same time, the rules of exponents say the order is commutative so we should not have two different answers. How do we reconcile this?
Neither, to calculate $x^{2*1/2}$ you do the multiplication first to get $x^1=x$.
And
$\sqrt{x}^2=(x^{1/2})^2$
$\sqrt{x^2}=(x^2)^{1/2}$
but in general, $x^{ab}\ne (x^a)^b$
First see the comments of jjagmath and myself that follow this answer.

The rule of exponents does say that order is commutative, so you never have two different answers.  However, the domain of the two functions is different.  Therefore, an element $x$ can be in the domain of one of the functions, and out of the domain of the other function.
Personally, I think the expression $x^{2 \times (1/2)}$ is ambiguous and might reasonably be construed to represent either

$(x^2)^{(1/2)}$ or
$x^{2 \times (1/2)} = x^{1}$, since $[2 \times (1/2)] = 1.$

It depends on the order of evaluation.
Personally, I would not evaluate it as $\left(\sqrt{x}\right)^2$, because I regard that as being represented by
$x^{(1/2) \times 2}$, which I regard as different from
$x^{2 \times (1/2)}$.

Note that $\sqrt{x^2} = |x|$
and that (in Real Analysis) 
$\left(\sqrt{x}~\right)^2$ is undefined when $x &lt; 0.$  
When $x \geq 0$, then $\left(\sqrt{x}~\right)^2$ evaluates the same way that $\sqrt{x^2}$ does.