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? [closed]

$1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6} , 7^{1/7}$.

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


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 $0< x<e$ and decreasing for $e<x$.

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.

2^{1/2} &\stackrel{?}{=} 3^{1/3} \\
\left(2^{1/2}\right)^6 &\stackrel{?}{=} \left(3^{1/3}\right)^6 \\
2^3 &\stackrel{?}{=} 3^2 \\
8 &\stackrel{?}{=} 9

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


Source : Link , Question Author : anshabhi , Answer Author : chharvey

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