# Which of these numbers is greater: $\sqrt[5]{5}$ or $\sqrt[4]{4}$?

I know that this is the question of elementary mathematics but how to logically check which of these numbers is greater: $\sqrt[5]{5}$ or $\sqrt[4]{4}$?

It seems to me that since number $5$ is greater than $4$ and we denote $\sqrt[5]{5}$ as $x$ and $\sqrt[4]{4}$ as $y$ then $x^5 > y^4$.

• The LaTeX commands work if you insert a dollar sign on each side of the mathematical expression in question. I've done that for you; I hope that's OK ... – Amitesh Datta Jul 26 '13 at 9:55
• Which one is greater? $(\sqrt[5]{5})^{n}$ or $(\sqrt[4]{4})^{n}$? (Find a suitable $n$.) – egreg Jul 26 '13 at 9:56

$\text{}$$5^4<4^5$$\text{}$
If $x_0$ is some positive natural number (or in fact any real number greater than $\tfrac{1}{\text e}$), then $$\left(\frac{\text d}{\text dx}x^x\right)_{x=x_0}=\left(x^x(\ln(x)+1)\right)_{x=x_0}>0.$$ The function is smooth and growing, so bigger numbers $x$ give bigger $x^x$.
• Yes, that's true but we're not looking at numbers of the form $x^{x}$ here; rather, we're looking at numbers of the form $x^{\frac{1}{x}}$ ... – Amitesh Datta Jul 26 '13 at 10:03