$\sqrt[x]{y} = y^{1 / x}$? Need help with algebra of exponentiation I need help with algebra of exponentiation.
 
$$\begin{align*}
\sqrt{x^2} &= (x^2)^{1/2} &\qquad &\text{(since }\sqrt[x]{y}=y^{1/x}\text{)}\\
&= x^{2(1/2)} &&\text{(since }(x^y)^z = x^{yz}\text{)}\\
&= x^{(1/2)2} &&\text{(since }xy=yx\text{)}\\
&= (x^{1/2})^2 &&\text{(since }(x^y)^z = x^{yz}\text{)}\\
&= \left(\sqrt{x}\right)^2 &&\text{(since }\sqrt[x]{y} = y^{1/x}\text{)}
\end{align*}$$

$\sqrt{x^2}$ is a real number.
$(\sqrt{x})^2$ is a real number, if $x\geq 0$. 
$\sqrt{x^2} = (\sqrt{x})^2$. 

The equality
$$
(x^2 )^{1/2}  = x^{2(1/2)} 
$$
holds only for $x \geq 0$. If $x < 0$, then 
$$
(x^2 )^{1/2} = |x| > 0,
$$ 
but
$$
x^{2(1/2)} = x < 0.
$$
