Order of exponents matters?

Question

I always thought that $(a^b)^c=a^{bc}$...

I am confused about why the order of exponents seems to matter in this particular case:

$((x-0.5)^2)^{1/3}$

is not the same as

$(x-0.5)^{2/3}$

However if I take the norm of the equation then they are equal.

Answer 1

The handling of $x^{2/3}$ depends on the grapher. If the grapher handles only real numbers and is smart enough, or if it handles either reals or complex numbers and has a setting to handle just reals and it is smart enough, it can handle $x^{2/3}$ as $\sqrt[3]{x^2}$, which means it can handle all values of $x$.

However, some graphers are not that smart and they cannot handle negative values of $x$. Some try to do it as $e^{\frac 23\ln x}$ which cannot handle negative $x$, or for that matter zero $x$. Some try other methods but still fail for negative $x$. An example of this is the TI-84 graphing calculator with older operating systems.

On the other hand, some can handle complex numbers. If the setting is to allow complex numbers, fractional powers can unexpectedly result in complex results. For example, in real numbers. $(-1)^{2/3}=1$, but in complex numbers,

$$(-1)^{2/3}=-\frac 12+\frac{\sqrt 3}2i$$

In this case, the grapher cannot handle $x^{2/3}$ as expected. An example is the current TI-Nspire CX calculator. In real settings, it graphs your function as expected. When set to handle complex numbers, it acts in the unexpected way. Here is a comparison, first in real mode:

And here is "rectangular" complex mode:

This caused me trouble in my calculus class last week. My calculator would not graph as expected in front of my students: how embarrassing! The cause was the complex setting. Resetting back to real mode solved the problem. But of course I did not realize that until class was over.

Answer 2

The problem is that raising a negative number to a noninteger power is multivalued and complex. For example, $(-1)^{1/2}$ can equal both $i$ and $-i$. The computer program you are using doesn't know which of the values of $(x-.5)^{2/3}$ you want to see for $x<.5$, so it shows you nothing.