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What is the larger of the two numbers?

$$\sqrt{2}^{\sqrt{3}} \mbox{ or } \sqrt{3}^{\sqrt{2}}\, \, \; ?$$ I solved this, and I think that is an interesting elementary problem. I want different points of view and solutions. Thanks!

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    $\begingroup$ This again $e^\pi$ or $\pi^e$ $\endgroup$
    – checkmath
    Commented May 22, 2012 at 3:14
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    $\begingroup$ @Gerry, although I liked your comment, but I might disagree that all we want is to get our answer accepted. First, we can still get likes form others if the answer is correct (see for example Robert's answer). Second, we need to get those questions answered for the benefits of others anyway. $\endgroup$
    – Rafid
    Commented May 22, 2012 at 5:33
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    $\begingroup$ @Rafid: Actually, it is not about people getting their answer accepted, but about askers such as this one acknowledging the help they've received. It is the polite thing to do! $\endgroup$
    – user641
    Commented May 22, 2012 at 8:16

6 Answers 6

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$$\sqrt2^{\sqrt 3}<^?\sqrt3^{\sqrt 2}$$ Raise both sides to the power $2\sqrt 2$, and get an equivalent problem: $$2^{\sqrt 6}<^?9$$ Since $\sqrt 6<3$, we have: $$2^{\sqrt 6}< 2^3 = 8 <9$$ So ${\sqrt 2}^{\sqrt 3}$ is smaller than $\sqrt3^{\sqrt 2}$.

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Hint: If $a$ and $b$ are positive numbers, $a^b < b^a$ if and only if $\dfrac{\ln a}{a} < \dfrac{\ln b}{b}$. Find intervals on which $\dfrac{\ln x}{x}$ is increasing or decreasing.

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  • $\begingroup$ Yet 2<e<3, so how? $\endgroup$
    – Wok
    Commented May 22, 2012 at 9:03
  • $\begingroup$ It answers the question for $e^\pi$ or $\pi^e$. $\endgroup$
    – Wok
    Commented May 22, 2012 at 9:08
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    $\begingroup$ @wok: try $\sqrt2$ and $\sqrt3$ (not $2$ and $3$). $\endgroup$
    – Did
    Commented May 22, 2012 at 11:54
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    $\begingroup$ @Didier Ok, my mistake. $\endgroup$
    – Wok
    Commented May 22, 2012 at 12:17
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We have $\sqrt{2}>1$ and $\sqrt{3}>1$, so raising either of these to powers $>1$ makes them larger.

Call $x=\sqrt{2}^\sqrt{3}$ and $y=\sqrt{3}^\sqrt{2}$.

We have $x^{2\sqrt{3}}$=8 and $y^{2\sqrt{2}}=9.$

Since $2\sqrt{2} < 2\sqrt{3}$, we conclude $y>x$.

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Hint: Use the Logarithm function.

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    $\begingroup$ I downvoted as it was not useful. Might as well say "potato" unless one assumes one already knows how to do it, defeating the purpose. $\endgroup$ Commented May 22, 2012 at 3:27
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    $\begingroup$ You can add a similar idea to Robert's, if that is what you had in mind. $\endgroup$
    – Pedro
    Commented May 22, 2012 at 4:56
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    $\begingroup$ This is a fine hint. The asker may have not thought at all about the logarithm function. $\endgroup$
    – Potato
    Commented Jul 21, 2012 at 5:44
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    $\begingroup$ You're just saying that because your name is Potato. $\endgroup$ Commented Oct 18, 2012 at 4:31
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In general, we can state two pertinent results: (1) If $a$ and $b$ are positive real numbers such that $b > a \ge e,$, then $a ^ {b} > b ^ {a}$; (2) If a and b satisfy $e \ge b > a > 0$, then $b ^ {a} > a ^ {b}.$

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$$\sqrt{2}^{\sqrt{3}} \approx 1.414^{1.732} \approx 1.822$$ $$\sqrt{3}^{\sqrt{2}} \approx 1.732^{1.414} \approx 2.174$$ $$\text{The rest is clear.}$$

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    $\begingroup$ I assume the problem is meant to be solved without a calculator, which may not be easy from your solution. $\endgroup$
    – Joe
    Commented May 23, 2012 at 3:30
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    $\begingroup$ hahah nice one! $\endgroup$
    – lesnikow
    Commented May 23, 2012 at 4:42

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