Why does $x^{\frac{3}{4}}$ have a domain of $x≥0$, while $x^{\frac{6}{8}}$ has a domain of $x\in{\Bbb{R}}$?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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