Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$ Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sqrt[60]{4^{15}}=\sqrt[60]{2^{30}}\\\sqrt[5]{5}=\sqrt[60]{5^{12}}\\\sqrt[6]{6}=\sqrt[60]{6^{10}}=\sqrt[60]{2^{10}\cdot3^{10}}$$ Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$? I can't come up with another approach.

The following fills-in the remaining step in OP's approach.

Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$?


*

*$3^{20} = 9^{10} \gt 8^{10}=2^{30}\,$ which excludes $\,\sqrt{2} = \sqrt[4]{4}\,$ as possible maximums;


*$3^{20} = 9^{10} \gt 6^{10}$ which excludes $\,\sqrt[6]{6}\,$ as a possible maximum;


*$3^{20} \gt 3^{18} = 27^6 \gt 25^{6} = 5^{12}$ which excludes $\,\sqrt[5]{5}\,$ as a possible maximum.
