Find $x$ such that $\sqrt[x]9=81$ Find $x$ such that $\sqrt[x]9=81$
If I simplified this to $9^{\frac1x}=81$, then we have $x={\frac12}$
I'm stuck here,
My question:  can we rewrite $\sqrt[\frac{a}{b}] c\quad$ as $\sqrt[a]{c^b}\quad$ for all positive  real number $c$?

Let $\,a,b\in\mathbb N\;$ and let $\,c\in\mathbb R^+\,.$
We could define the symbol $\,\sqrt[\frac ab]c\,$ as that positive real number $\,x\,$ (it is unique) such that $\,x^{\frac ab}=c\,.$
Therefore ,
$x^a=c^b\;,$
$x=\sqrt[a]{c^b}\;.$
Hence ,
$\sqrt[\frac ab]c=x=\sqrt[a]{c^b}\,.$
