Find $x$ such that $\sqrt[x]9=81$

Question

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 $c$?

Answer 1

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 ,

$\sqrt[\frac ab]c=x=\sqrt[a]{c^b}\,.$

Answer 2

Yes you can! Note that $$\sqrt[\frac a b]c=c^{\frac {1} {\frac a b}}=c^{\frac b a}=\sqrt[a]{c^b}$$