Which of the numbers $1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6} , 7^{1/7}$ is largest, and how to find out without calculator? $1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6} , 7^{1/7}$.
I got this question in an Application of Derivatives test. I think log might be used here to compare the values, but even then the values are very close to each other and differ by less than 0.02, which makes it difficult to get some specific answer to this question. How to solve this by a definite method?
Source: ISI entrance exam

Since $1$ is obviously smaller than all the rest, we can skip it.
Check the rest by taking a pair of numbers, comparing them and proceeding with the larger one:



*

*Take $2^{1/2}$ and $3^{1/3}$

*Raise them both to the power of $6$ (the LCM of $2$ and $3$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(2^{1/2}\right)^{6}=2^3=8<9=3^2=\left(3^{1/3}\right)^{6}$$



*

*Take $3^{1/3}$ and $4^{1/4}$

*Raise them both to the power of $12$ (the LCM of $3$ and $4$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(3^{1/3}\right)^{12}=3^4=81>64=4^3=\left(4^{1/4}\right)^{12}$$



*

*Take $3^{1/3}$ and $5^{1/5}$

*Raise them both to the power of $15$ (the LCM of $3$ and $5$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(3^{1/3}\right)^{15}=3^5=243>125=5^3=\left(5^{1/5}\right)^{15}$$



*

*Take $3^{1/3}$ and $6^{1/6}$

*Raise them both to the power of $6$ (the LCM of $3$ and $6$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(3^{1/3}\right)^{6}=3^2=9>6=6^1=\left(6^{1/6}\right)^{6}$$



*

*Take $3^{1/3}$ and $7^{1/7}$

*Raise them both to the power of $21$ (the LCM of $3$ and $7$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(3^{1/3}\right)^{21}=3^7=2187>343=7^3=\left(7^{1/7}\right)^{21}$$

Hence the answer is $3^{1/3}$.
