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

Consider the function $f(x) = x^{1/x}$. We ask, when is this maximized? So we ask, what is $f'(x)$? To do this, we use logarithmic derivatives.
$$ \log f(x) = \log x^{1/x} = \frac{\log x}{x},$$
so
$$ \frac{f'(x)}{f(x)} = \frac{1 - \log x}{x^2},$$
or rather
$$ f'(x) = \frac{1 - \log x}{x^2} x^{1/x}.$$
Notice that this is positive when $1 > \log x$ and negative when $1 < \log x$, so the actual max is at $x = e$. This means that the max value of $f(x)$ will be at either $2^{1/2}$ or $3^{1/3}$. We might guess that since $e$ is closer to $3$, that $e^{1/3}$ will be the bigger of the two. Using whatever method of your choosing to evaluate $2^{1/2}$ and $3^{1/3}$, we can confirm that heuristic: $3^{1/3}$ is larger.
