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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 $x^5 > y^4$.

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    $\begingroup$ 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 ... $\endgroup$
    – Amitesh Datta
    Commented Jul 26, 2013 at 9:55
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    $\begingroup$ Which one is greater? $(\sqrt[5]{5})^{n}$ or $(\sqrt[4]{4})^{n}$? (Find a suitable $n$.) $\endgroup$
    – egreg
    Commented Jul 26, 2013 at 9:56

3 Answers 3

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$\text{}$$5^4<4^5$$\text{}$

Now, take the 20th root on both sides of the inequality if you can!

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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 $x^x$.

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    $\begingroup$ 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 $x^{\frac{1}{x}}$ ... $\endgroup$
    – Amitesh Datta
    Commented Jul 26, 2013 at 10:03
  • $\begingroup$ @AmiteshDatta: Oh, yeah that's right. My bad. $\endgroup$
    – Nikolaj-K
    Commented Jul 26, 2013 at 10:41
  • $\begingroup$ No problem, Nick! I would still upvote your answer nonetheless (because it presents a new idea) but I've exhausted my daily vote limit of 40 votes. But I will upvote your answer tomorrow! $\endgroup$
    – Amitesh Datta
    Commented Jul 26, 2013 at 11:19
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    $\begingroup$ @AmiteshDatta: Don't worry, I'm not a fan of the reputation system anyway. (It makes people feel they need to write in a certain style once they got some, and on the other side, if a poster with many points posted an answer, even if it's only an OK answer, often no more answers follow. I recently spent 1500 points on bounties just to get rid of it. On the physics board, I spent about 5000 points on bounties - and I can now empirically say that it doesn't improve the answer quality. The questions which get bounties are usually so hard (or broad) that nobody bothers anyway.) $\endgroup$
    – Nikolaj-K
    Commented Jul 26, 2013 at 11:53
  • $\begingroup$ I'm sorry Nick but could you explain this to me better, or maybe draw a graph, I think I'm a little bit cofused $\endgroup$
    – mike
    Commented Jul 26, 2013 at 12:29
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The function $$n^{1/n}=e^{(1/n)\log n}$$ goes to $1$ as $n$ becomes infinite. Also, taking derivatives shows that it is monotonically decreasing whenever $n>e.$ If we consider only integers now, we have that for $n=4,5,6,\ldots,$ the sequence decreases.

It follows that $$4^{1/4}>5^{1/5}.$$

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