How does the exponent of a function effect the result?

The $x^{2/2}$ can be represented by these ways: $$\begin{align} x^{2\over2}=\sqrt{x^2} = |x|\\ \end{align} $$ And
$$\begin{align} x^{2\over2}=x^{1} = x\\ \end{align} $$ Which one is correct? And what is the domain of $x^{2 \over 2}$?

Solution 1:

$2/2 = 1$, so $x^{2/2} = x^1 = x$. Always. What is confusing you is that this may not be the same as $(x^2)^{1/2}$ or $(x^{1/2})^2$ when $x$ is not positive. In mathematics, an expression such as $x^p$ depends on the values of $x$ and $p$, not on the way they are represented.

Solution 2:

$(x^2)^{1/2}$ is defined for every real $x$, but $(x^{1/2})^2$ is defined for $ x \ge 0$

Solution 3:

The rule $$ x^{pq} = (x^p)^q $$ is not always valid if $p$ or $q$ are not integers and $x<0$. As you have noticed $$ -1 = (-1)^1 = (-1)^{\frac 2 2} \neq ((-1)^2)^{\frac 1 2} = 1^{\frac 1 2} = 1. $$

So, unfortunately, $x^{\frac 2 2} \neq \sqrt{x^2}$ for $x<0$. For the last question: $x^{\frac 2 2} = x^1 = x$ is defined for all $x$.

Solution 4:

I'm amazed to ctrl-f 'order of operations' and come up with nothing here!

Other answers do well to explaining what's going on here, but recognize that the confusion is about perceived ambiguity with respect to order of operations. In this case, there is an invisible (but understood) set of brackets in the exponent $x^{(\frac{2}{2})}$, the same way that $\frac{5+1}{2}$ is understood to be the same as $\frac{(5+1)}{2}$.

All of this is just convention, which of course is malleable (even in mathematics). Exercise care to express what you mean in a way that is unambiguous to yourself and to whoever you are trying to communicate with.