Why $x^{(1/2)2} \neq x^{2(1/2)} $? I know, probably is a newb. question, but i can't get this $x^{(1/2)2} \neq x^{2(1/2)} $ $ x\in\mathbb R^+$.
I know $x^{(1/2)2}=(\pm \sqrt{x})^2=+x $ 
and $x^{2(1/2)}=\pm x $ because $(+x)^2=(-x)^2=x^2$.
Edit so  $\pm x \neq +x$
so...maybe multiplication is not commutative and $(1/2)2 \neq 2(1/2)$, or maybe $\sqrt[n]{x}\neq x^{1/n}$... or $(x^n)^m\neq x^{n\times m} $ in other words it isn't a multiplication.
Maybe im wrong when I think that $\log_x(x^y)=y$ ($\log_x$ isn't multivalued for $x,y>0$ or not?) and 
$x^{(1/2)2} \neq x^{2(1/2)} \rightarrow (1/2)2 \neq 2(1/2)$.
What am I missing? 
 A: It depends on what your definitions are.
Usually, $f : x \mapsto x^2$ is a function from $\Bbb R$ to $\Bbb R^+$, and $g : x \mapsto x^{1/2}$ is a function from $\Bbb R^+$ to $\Bbb R^+$.
So, $f \circ g$ is defined on $\Bbb R^+$ and forall $x \in \Bbb R^+, f(g(x)) = f(\sqrt x) = \sqrt x \sqrt x = x$.
On the other hand, $g \circ f$ is defined on $\Bbb R$ and forall $x \in \Bbb R, g(f(x)) = g(x^2) = \sqrt{x^2} = |x|$.
$g \circ f$ is an extension of $f \circ g$ to $\Bbb R$.
So, if $x\ge 0$; it is true that $(x^{1/2})^2 = x^{(1/2)2} = x^1 = x^{2(1/2)} = (x^2)^{1/2}$.
If $x < 0$, then $(x^{1/2})^2$ doesn't make any sense but the other equalities are still valid. 
In general, the rule $x^{ab} = (x^a)^b = (x^b)^a$ is only valid for real positive $x$ and real exponents $a$ and $b$ ; or for real or complex $x$ and integers exponents $a$ and $b$. Using it blindly in other cases is dangerous.
A: You are wrong because 
$$x^{2 (1/2)}=x^1= x$$ 
Something related is that  in general 
$$(a^n)^m\neq a^{nm}$$
(when $a$ is negative for example, or when $n$ or $m$ are complex)
In general
$$\sqrt[n]{x^m}\neq x^{\frac{m}{n}}$$
and $$
\log(x^y) \neq y \log(x)$$
as $x<0$ 
