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?
You can use calculus to find where the maximum of $f(x)=x^{1/x}$ occurs. You will find that it is at $x=e$. You will also find that $f(x)$ is increasing for $0&lt; x&lt;e$ and decreasing for $e&lt;x$.

Unfortunately, the choice $e^{1/e}$ was not given. The two values of $x$ that you are given that bracket $e$ are $2$ and $3$, so the correct choice is either $2^{1/2}$ or $3^{1/3}$.

We can decide between the two of them by raising both sides to the sixth power.

\begin{align}
2^{1/2} &amp;\stackrel{?}{=} 3^{1/3} \\
\left(2^{1/2}\right)^6 &amp;\stackrel{?}{=} \left(3^{1/3}\right)^6 \\
2^3 &amp;\stackrel{?}{=} 3^2 \\
8 &amp;\stackrel{?}{=} 9
\end{align}

We see that $x=3$ gives us the larger function value, so the correct answer is

$$3^{1/3}$$
If you haven't been introduced to elementary calculus yet, you can take a look at this solution. Comparing integers is easier than comparing irrational numbers. Since $[1, 2, \cdots, 7] = 420$, we can take the $420$th powers of all these numbers will give us a bunch of integers to compare: $$1, 2^{210}, 3^{140}, 2^{210}, 5^{84}, 2^{70} 3^{70}, 7^{60}.$$ We see after some basic arithmetic that $3^{140}$ is the largest among these numbers. Taking the $140$th root, it is obvious that $3^{1/3}$ is the largest among the original bunch of numbers.