Under what conditions does $x^{\frac{b}{c}} = (x^b)^\frac{1}{c}$ hold?

Question

It is very common to use the formula $$x^{\frac{b}{c}} = (x^b)^\frac{1}{c}$$ to simplify the evaluation of a fractional exponent.

I want to know what circumstances allow us to do this step. For example, it does not work in this situation: $$(-4)^{\frac{2}{4}} = ((-4)^2)^\frac{1}{4} = 16^\frac{1}{4} = 2$$ The correct answer is $2i$, but the formula yields $2$. What caused it to go awry here, and in the general case, how can we avoid errors occurring for this reason?

Answer 1

By definition $x^{\alpha}$ is equal to $\textrm{exp}(\alpha \, \textrm{log} \, x)$. So by definition it is not defined if $x$ is not a in $\mathbb R^*_+$. As a consequence, you cannot say that $\sqrt{-4} = 2i$ is the correct answer.

All this depends on a choice you make at first : the choice to extend the logarithm function from $\mathbb R^*_+$ to $\mathbb C$. Any two choices are equal modulo $2i \pi$. In particular, for any $z \in \mathbb C$, there exists $n$ such that $\mathrm{log}(\mathrm{exp} \, z) = z + 2in\pi$.

Here are the details of what fails in your example : $ (x^b)^{\frac{1}{c}} = \mathrm{exp}(\frac{1}{c}\mathrm{log}(e^{b \mathrm{log} \,x}))$, so there exists $n$ such that $\mathrm{log}(e^{b \mathrm{log} \, x}) = b \mathrm{log} \, x + 2in\pi$. Then $(x^b)^{\frac{1}{c}} = x^{\frac{b}{c}} \times \mathrm{exp}(\frac{2in\pi}{c})$.

The conclusion is that the formula works only if $x$ is in $\mathbb R^*_+$.

Answer 2

It holds always.

But De Moivre's Theorem says that $z^{\frac{1}{q}}$ is one of the values from the set $\{|z|^{\frac{1}{q}}\xi: \xi^q=1\}$, provided $q\in\mathbb{Z}^+$.

So, $16^\frac14\in\{2,-2,2i,-2i\}$

Answer 3

Correct way to say, would be:

Suppose we are given this equation to solve over $\mathbb{C}$;
$\displaystyle x^q=x_o^p$ where $x_o$ is a given complex number and $p,q\in\mathbb{Z}$ are coprime integers.

Then $z$ is simply what I stated before.

But if $\gcd(p,q)=d>1$ then it forces another condition that $x^\frac{p}{d}=x_0^{\frac{q}{d}}$. So, again by the above method we can find such solution.

For example- But we have $x=(-4)^\frac24$.
This constraints the value of $x$ satisfying both $x^4=16$ as well as $x^2=-4$.

As $x$ satisfying the latter would obviously satisfy the former we just solve $x^2=-4$ to get $x\in\{2i,-2i\}$

Answer 4

The "precalculus" answer would be: it holds when $x>0$. And maybe provide an example (like the one given) that it can fail when $x<0$ using a "precalculus" understanding of the symbols. Later, when complex analysis is studied, the more intricate nature of $x^y$ can be considered.