Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
It will be true for $x>0$ since we can apply logarithms. I'm not sure, but I'd guess this is precisely why the example you give goes awry. –  Clayton Jan 24 '13 at 3:22
I think the definition of $x^{b/c}$, when $x<0$, has to be made under the assumption that $\gcd(b,c)=1$, or at least that $b,c$ are not both even. With that assumed, the definition as the $c^{th}$ root of $x^b$ seems to work, where when the $c$ is even and $x<0$, the fractional expression $x^{b/c}$ is considered as undefined. –  coffeemath Jan 24 '13 at 3:28
add comment

4 Answers

up vote 2 down vote accepted

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 $\mathbb R^*_+$.

share|improve this answer
add comment

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

share|improve this answer
I'm not sure I understand. Are you saying that $16^\frac{1}{4}$ is $-2$ in this case? –  Peter Olson Jan 25 '13 at 3:09
I mean that all the 4th roots of $16$ are $2,-2,2i,-2i$. So you can't be sure of which one to choose as all satisfy $x^4=16$. –  Grobber Jan 25 '13 at 16:16
I know that there are four solutions to $x^4 = 16$, but I was pretty sure that $16^\frac{1}{4}$ always returned the same single value, similar to how $(x^2)^\frac{1}{2}$ always returns $|x|$. If I calculate $16^\frac{1}{4}$, I always get 2, never -2, 2i, or -2i. –  Peter Olson Jan 26 '13 at 15:03
I think its for our brevity that we mean $\sqrt{x^2}=|x|$. –  Grobber Jan 27 '13 at 9:07
I think we pick it so that $f(x) = \sqrt{x}$ can be a function. –  Peter Olson Jan 27 '13 at 13:57
show 2 more comments

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.

share|improve this answer
add comment

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 $x\in\{2i,-2i\}$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.