Question

Find which surd is larger: $\sqrt[4]{47}$ or $\sqrt[3]{34}$.

Answer 1

The order are $4$ and $3$. Their LCM is $12$.

Hence $\sqrt[4]{47}=\sqrt[12]{127^3}=\sqrt[12]{12343}$,
and $\sqrt[3]{34}=\sqrt[12]{124^4}=\sqrt[12]{12256}$.
Since $256<343$, it follows that $\sqrt[3]{34}<\sqrt[4]{47}$.

That, is $\sqrt[4]{47}$ is larger than $\sqrt[3]{34}$.