Why is $({x}^{3/2})^{2/3} = x$ while $({x}^{2/3})^{3/2} = |x|$?

Question

Here's my thinking: $$\sqrt[3]{{\sqrt[2]{x}^3}}^2$$ $$\sqrt{x}^2$$ $$|x|$$ What am I missing? Some calculators just multiply the fractions in both cases and give $x^1$, but it seems that the answer is |x| at least for the second case.

Answer 1

The exponential function $z^w$ is only well-defined for real, positive $z$ or integer $w$. When $z$ is not a positive real number and $w$ is not an integer, you can get these sorts of inconsistencies depending on how exactly the exponential is being handled.

One convention, usually used when $z$ is known to be real, is to always use the real root if it exists, and consider the expression undefined if it does not. If this convention is used, then $$ (x^{2/3})^{3/2} = |x|\;\;\;;\;\;\;(x^{3/2})^{2/3} = \begin{cases} x & x \ge 0 \\ \mathrm{undefined} & x \le 0\end{cases}. $$

Another convention that allows for $z$ to be complex (and is never undefined) is to define in terms of the exponential and natural logarithm functions: $$ z^w = e^{w \,\mathrm{Ln}(z)} = |z|^we^{iw\mathrm{Arg}(z)}. $$ Under this convention, $(x^{2/3})^{3/2} = x$ but $$ \left(x^{3/2}\right)^{2/3} = \exp\left[\frac{2}{3}\mathrm{Ln}\left(|x|^{3/2}e^{3i\mathrm{Arg}(z)/2}\right)\right] = x\omega $$ where $\omega = 1$ if $\mathrm{Re}[x]/|x| < -1/2$ and $[-1+ i\, \mathrm{sgn}(\mathrm{Im}[x])\sqrt{3}]/2$ otherwise.

Both of these conventions agree that $(x^{2/3})^{3/2} = (x^{3/2})^{2/3} = x$ when $x \ge 0$. But things go awry when we allow $x$ to be negative or complex.

Answer 2

$x^{3/2}$ or $x^{2/3}$ makes no sense for $x<0$.

$x^r$ for $r\notin\mathbb Z$ is defined for $x>0$ (and $x=0$ if $r\ge0$).

Answer 3

The equality $(a^b)^c=a^{bc}$ holds when $a\geqslant0$, but if $a<0$ it may not make sense. For instance, what is $(-1)^{1/2}$? It is undefined (in $\mathbb R$) and therefore it makes no sense to assert that $\left((-1)^{1/2}\right)^2=(-1)^1=-1$.

Answer 4

Before starting to simplify the expression, you have to find the condition of existence of the square root. $\sqrt{x}$ can be calculed only if $x\geq 0$. This is the reason of $|x|$.