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?

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    $\begingroup$ 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. $\endgroup$ – Clayton Jan 24 '13 at 3:22
  • $\begingroup$ 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. $\endgroup$ – coffeemath Jan 24 '13 at 3:28

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 $\mathbb R^*_+$.


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

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    $\begingroup$ I'm not sure I understand. Are you saying that $16^\frac{1}{4}$ is $-2$ in this case? $\endgroup$ – Peter Olson Jan 25 '13 at 3:09
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    $\begingroup$ 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 $x^4=16$. $\endgroup$ – Grobber Jan 25 '13 at 16:16
  • $\begingroup$ 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. $\endgroup$ – Peter Olson Jan 26 '13 at 15:03
  • $\begingroup$ I think its for our brevity that we mean $\sqrt{x^2}=|x|$. $\endgroup$ – Grobber Jan 27 '13 at 9:07
  • $\begingroup$ I think we pick it so that $f(x) = \sqrt{x}$ can be a function. $\endgroup$ – Peter Olson Jan 27 '13 at 13:57

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.


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


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