Question about whether $(x^2)^{0.5} = x$. The Wikipedia page on exponentiation suggests that the following identity holds provided the base $b$ is non-zero:
$$(b^m)^n = b^{mn}$$
Consider the following function:
$$y = (x^2)^{0.5}$$
According to the identity above the following should hold:
$$y = (x^2)^{0.5} = x^1$$
However consider the plots of the two functions:
$$y = x$$
$$y = (x^2)^{0.5}$$
The functions are equal for $x \geq 0$ however for $x < 0$ there is a discrepancy. Could you comment on the use of exponentiation rules? Are there other circumstances that similar discrepancies can be found when applying exponentiation rules?
 A: The page you refer to actually says the following.  (The emphasis below is mine.)

The following identities hold for all integer exponents, provided that the base is non-zero

$0.5$ is not an integer, so the property does not apply.
But as you've seen, $(x^2)^{0.5} = x$ is true for $x \ge 0$.  The reason it fails for $x < 0$ is because raising to the exponent $0.5$ is the same as taking the square root (as long as we're working only with real numbers, which I'm assuming is the case here).  That is, $y^{0.5} = \sqrt{y}$.  And when we simply say $\sqrt y$, it is understood that we are talking about the positive square root.  For example, $\sqrt9 = 3$, not $-3$.
In general when we take the square root of a square, we have $\sqrt{x^2} = |x|$.  For example, if $x = -4$ then we have $$x^{0.5} = ((-4)^2)^{0.5} = (16)^{0.5} = 4 \ne x.$$
A: The identity you stated only holds for $x \geq 0$ (as noted in your wikipeda link) so, for general $x \in \mathbb{R}$ we have $(x^2)^{1/2} = |x| \neq x$ for $x <0$, hence the discrepancy 

A clarification; you have two options:


*

*Either the identity holds for non-zero bases provided the exponents are integers 

*The identity holds for real exponents provided the base is non-negative. 
A: Traditionally, exponentiation of real numbers is ill defined when the base is negative and the exponent is not an integer. That is because of the problem you yourself just found.
A rational number can be represented in multiple ways, and the denominator can be odd or even depending on your choice of representation. If, say, you defined that $x^{\frac{a}{b}}=\sqrt[b]{{\left(x^a\right)}}$, and $x$ is negative, the parity of $a$ changes the sign of what's inside the radical.
For irrational numbers, things get even more confusing. When the basis is positive, one can extend exponentiation via continuity (since these sign problems don't exist); after all, any irrational can be approximated by rationals. The thing is there's no 'canonical approximation' and even if there were, why should it in principle be such that the numerators of the approximations are all the same parity?
When you dive into complex analysis, these problems are dealt with via logarithm branches. Even then, it still essentially means that there is no 'unique' or best way to do so.
