$\sqrt[x]{y} = y^{1 / x}$? Need help with algebra of exponentiation

Question

I need help with algebra of exponentiation.

$$\begin{align*} \sqrt{x^2} &= (x^2)^{1/2} &\qquad &\text{(since }\sqrt[x]{y}=y^{1/x}\text{)}\\ &= x^{2(1/2)} &&\text{(since }(x^y)^z = x^{yz}\text{)}\\ &= x^{(1/2)2} &&\text{(since }xy=yx\text{)}\\ &= (x^{1/2})^2 &&\text{(since }(x^y)^z = x^{yz}\text{)}\\ &= \left(\sqrt{x}\right)^2 &&\text{(since }\sqrt[x]{y} = y^{1/x}\text{)} \end{align*}$$

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$\sqrt{x^2}$ is a real number.

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

$\sqrt{x^2} = (\sqrt{x})^2$.

Answer 1

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 $$ x^{2(1/2)} = x < 0. $$

Answer 2

With real numbers, the exponent rule $(x^y)^z = x^{yz}$ only holds for nonnegative $x$; it does not hold for arbitrary $x$. So you cannot go from line 1 to line 2 (in the edited version) except if you know $x\geq 0$. If you don't know whether $x\geq 0$ or not, then the equality $(x^y)^z = x^{yz}$ need not hold.

For example: $\sqrt{(-1)^2}$ is not equal to $(-1)^{(1/2)2}$.

In short,
$$\large(x^y)^z = x^{yz}$$ is not universally true; for integer exponents it holds for any base, but for nonintegral exponents it need not hold with nonpositive bases. In particular, for $y=\frac{1}{2}$, it can only hold if $x\geq 0$.