Question

Arrange $\sqrt[4]{4\sqrt[5]{58}},\sqrt[6]{6\sqrt[4]{416}}$ and $\sqrt[5]{5\sqrt[6]{624}}$ in ascending order

Answer 1

Write these in index form:
$\sqrt[4]{4\sqrt[5]{58}}={\left(\right(8{)}^{1/5})}^{1/4}=(8{)}^{1/20}$,
$\sqrt[6]{6\sqrt[4]{416}}={\left(\right(16{)}^{1/4})}^{1/6}=(16{)}^{1/24}$,
$\sqrt[5]{5\sqrt[6]{624}}={\left(\right(24{)}^{1/6})}^{1/5}=(24{)}^{1/30}$.
We observe the $LCM(20,24,30)=120$. Thus we can bring all of these two common index:
$\sqrt[4]{4\sqrt[5]{58}}=(8{)}^{1/20}=(8^6{)}^{1/120}$,
$\sqrt[6]{6\sqrt[4]{416}}=(16{)}^{1/24}=({16}^5{)}^{1/120}$,
$\sqrt[5]{5\sqrt[6]{624}}=(24{)}^{1/30}=({24}^4{)}^{1/120}$.
Now all the three have the same indices. It is sufficient to compare radicands $8^6,{16}^5$ and ${24}^4$. We further reduce them to
$8^6=2^{18},{16}^5=2^{20}$ and ${24}^4=3^4\times 8^4=3^4\times 2^{12}$.
All the three numbers have powers of two:
$2^{18}=2^{12}\times 2^6,2^{20}=2^{12}\times 2^8,3^4\times 2^{12}$.
Taking out $2^{12}$ from these numbers, it is enough to compare
$2^6=64,2^8=256$, and $3^4=81$.
Since $64<81<256$, we conclude that
$\sqrt[4]{4\sqrt[5]{58}}<\sqrt[5]{5\sqrt[6]{624}}<\sqrt[6]{6\sqrt[4]{416}}$.