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]

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.

\begin{align} 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 \end{align}

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

$$3^{1/3}$$

Hint: look at $f(x) = x^{1/x}$ and decide if this function is increasing or not. Write it as $x^{1/x} = e^{(\ln x)/x}$ to differentiate it easier.

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 $3^{1/3}$.