If $x^{1/2}$ is the same as $ \sqrt[2] x$ then why $x^{1/3}$ is not equal to $\sqrt[3] x$?

Question

If $x^{1/2}$ is the same as $\sqrt[2] x$ than why $x^{1/3}$ is not equal to $\sqrt[3] x$ and $x^{1/4}$ not equal to $\sqrt[4] x$ and so on...?

Thank you.

Answer 1

$x^{1/2}$ is just $\sqrt[2]{x}$ since :

$$ x^{1/2}x^{1/2}=x^{1/2+1/2}=x=\sqrt[2]{x}\sqrt[2]{x}, \tag*{$x\geqslant0$} $$

hence $x^{1/2}=\sqrt[2]{x}$.

Using the same idea you can show that $x^{1/n}=\sqrt[n]{x}$, for $n=2,3,\ldots$

Answer 2

It mostly depends on conventions. According to the one I follow, the function $x\mapsto x^{1/3}$ is only defined for $x>0$, because I use the definition $$ a^b=e^{b\log a} $$ when $b$ is not integer. Since $\log a$ is defined only for $a>0$, the map $x\mapsto x^{1/3}$ is defined only for $x>0$.

In this way I can say without posing any condition that, for instance, $$ x^{1/3}=x^{2/6}=(x^2)^{1/6} $$ which would be utterly false when $x<0$ and I defined $x^{1/3}=\sqrt[3]{x}$.

Other people accept $x^{1/3}$ also for negative $x$, but, in this case, algebraic manipulations like the one above require care.