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.
 A: Let$$f(x)=x^{1/x}=e^{\log(x)/x}$$and note that $f(n)=\sqrt[n]n$, for each $n\in\Bbb N$. You have$$f'(x)=\frac{1-\log(x)}{x^2}e^{\log(x)/x},$$which is greater than $0$ on $[1,e)$ and smaller than $0$ on $(e,\infty]$. Therefore $f$ is strictly increasing on $[1,e]$ and strictly decreasing on $[e,\infty)$. So, since $e<3$ and since $3<4<5<6$,$$\sqrt[3]3>\sqrt[4]4>\sqrt[5]5>\sqrt[6]6.$$Besides, $\sqrt2=\sqrt[4]4$. And it is easy to compare $\sqrt2$ with $\sqrt[3]3$; just use the fact that $\sqrt2^6=8$ and that $\sqrt3^6=9$.
A: 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.
A: You can do pairwise comparisons. For instance, to compare $\sqrt[4]4$ with $\sqrt[5]5$, you only need to compute $4^5$ and $5^4$. (And you can rule out $\sqrt[6]6$ easily by comparing it with $\sqrt[3]3$.)
A: It is also possible to work with logarithms.  We may write for each of these numbers
$$ \log_2  \sqrt2 \ \ = \ \ \log_2  2^{1/2} \ \ = \ \ \frac12 \ \ , \ \ \log_3  \sqrt[3]3 \ \ = \ \ \frac13 \ \ , \ldots , \ \ \log_6  \sqrt[6]6 \ \ = \ \ \frac{1}{6} \ \ .  $$
It will be more helpful to our purpose to express  these in terms of a common logarithmic base, say, $ \  \log_2 x  \ \ : $
$$ \log_2  \sqrt2 \ \  = \ \ \frac12 \ \ , \ \ \log_2  \sqrt[3]3 \ \ = \ \ \frac{\log_3  \sqrt[3]3}{\log_3 \ 2} \ \ = \ \ \frac{1}{3 \ \log_3 \ 2} \ \ ,   $$ $$ \log_2  \sqrt[4]4 \ \ = \ \  \frac{1}{4 \ \log_4 \ 2} \ \ , \ \ \log_2  \sqrt[5]5 \ \ = \ \  \frac{1}{5 \ \log_5 \ 2} \ \ , \ \ \log_2  \sqrt[6]6 \ \ = \ \  \frac{1}{6 \ \log_6 \ 2} \ \ . $$
We have one evident equality, $ \ \large{\frac{1}{4 \ \log_4 \ 2} \ = \ \frac{1}{4 \ \log_4 \ [4^{1/2}]} \ = \ \frac{1}{4 \ · \ (1/2)}   \ = \ \frac{1}{2} } \ \ . $
For the rest, we need to examine the consequences of some inequalities.  (All of the denominators in the ratios we've found are positive, so there are no undesired complications.)
$$ 2^3 \ < \ 3^2 \ \ \Rightarrow \ \ 2 \ < \ 3^{2/3} \ \ \Rightarrow \ \ \log_3 2 \ < \ \frac23 \ \ \Rightarrow \ \ \frac12 \ < \ \frac{1}{3 \ \log_3 2} \ \ ;  $$
$$ 2^5 \ > \ 5^2 \ \ \Rightarrow \ \ 2 \ > \ 5^{2/5} \ \ \Rightarrow \ \ \log_5 2 \ > \ \frac25 \ \ \Rightarrow \ \ \frac12 \ > \ \frac{1}{5 \ \log_5 2} \ \ ;  $$
$$ 5^6 \ = \ 15625 \ > \ 6^5 \ = \ 7776 \ \ \Rightarrow \ \ 5 \ > \ 6^{5/6} \ \ \Rightarrow \ \ \log_6 5 \ = \ \frac{\log_6 2}{\log_5 2} \ > \ \frac56 $$ $$  \Rightarrow \ \ \frac{1}{5 \ \log_5 2} \ > \ \frac{1}{6 \ \log_6 2} \ \ .  $$
Since $ \ \log_2 x \ $ is an increasing function of $ \ x \ $ on its domain $ \ x \ > \ 0 \ \ , \ $ the chain of inequalities
$$ \log_2  \sqrt[3]3 \ \ > \ \ \log_2  \sqrt2 \  =  \ \log_2  \sqrt[4]4 \ \ > \ \ \log_2  \sqrt[5]5 \ \ > \ \ \log_2  \sqrt[6]6    $$
implies
$$   \sqrt[3]3 \ \ > \ \    \sqrt2 \  =  \   \sqrt[4]4 \ \ > \ \   \sqrt[5]5 \ \ > \ \    \sqrt[6]6 \ \ .   $$
