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Why does the simplified version of $x^{\frac{6}{8}}$ have a domain of $x≥0$ while the unsimplified has a domain of $x\in{\Bbb{R}}$? Shouldn't they have the same domain being that they are the same expression? If a negative number were to be plugged into $x^{\frac{3}{4}}$ then it would take the 4th root of a negative number thus yielding imaginary numbers. But an unsimplified version of the same expression ($x^{\frac{6}{8}}$) would yield a positive number as it would take the 8th root of a positive number.

stoic-santiago
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Max Heppelmann
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  • Those are the exact same function, those are not two different objects, it's just one. So that one function has only one domain. – jjagmath Mar 25 '24 at 04:01

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The idea is, index law only can be happily applied in $\mathbb{R}^+$. In general (even in $\mathbb{C}$), we define $$x^y=\exp(y\log x)$$ and $\log$ is a multi-valued function in $\mathbb{C}$. Hence, most index law does not apply as usual outside $\mathbb{R}^+$.

In short, $x^\frac{2}{3}$ and $x^\frac{4}{6}$ are the same, but $x^\frac{4}{6}$ does not equal to $(x^4)^\frac{1}{6}$. And to prevent this kind of uncertainty, we would use the convention that $x^\frac{kp}{kq}$ to be the value of $x^\frac{p}{q}$, where $p,q$ are coprime.

Angae MT
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They should both have the same domain because both exponents are, in fact, equal. The domain stated here for the un-simplified faction is (I believe) wrong.

$\frac{6}{8}=\frac{3}{4}$ and therefore $x^{\frac{6}{8}}=x^{\frac{3}{4}}$.

I know some textbooks say differently on this, but I know of one (Cambridge University Press, Australian version) that fixed this in a more recent edition.

Red Five
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You see, they treat ${x}^{\frac{3}{4}}$ as $({x}^{3})^{\frac{1}{4}}$

The domain of ${x}^{3}$ is $x\in{\Bbb{R}}$

But the domain of ${x}^{\frac{1}{4}}$ is ${x}≥{0}$. Hence it seems to have a domain of (0,$∞$) This is how few websites like desmos, treat such functions. This is not a feasible method though, as their is no specific priority of solving fractional exponents.

In ${x}^{\frac{6}{8}}$, as ${x}^{6}$ is positive for all real numbers, $({x}^{6})^{\frac{1}{8}}$ has a domain of $x\in{\Bbb{R}}$. Hope this helps

Gwen
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  • Thanks @Gwen. I have a question for you (I know the answer, but I think others here may benefit from the discussion). Why do we interpret $x^{\frac{3}{4}}$ as $(x^{3})^{\frac{1}{4}} $ and not as $(x^{\frac{1}{4}})^{3}$? – Red Five Mar 25 '24 at 07:19
  • I am trying my best to explain it, I'm still in highschool :") According to me, any function which equates to ${x}^{y}$ can be represented as ${e}^{yln x}$ . In the given question,y is a rational number and can be represented by $\frac{a}{b}$ where b $≠$0. As natural logarithm of negative numbers is not defined, x should be positive. And hence the domain must be positive as well, not even zero. Actually , you won't be able to plot a graph of ${x}^{y}$ if $x<0$ in Cartesian plane as well. Maybe websites provide priority to the numerator, here 3, before the denominator. (Contd in next reply) – Gwen Mar 25 '24 at 10:50