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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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This again $e^\pi$ or $\pi^e$ –  checkmath May 22 '12 at 3:14
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@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. –  Rafid May 22 '12 at 5:33
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@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! –  user641 May 22 '12 at 8:16
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6 Answers

up vote 64 down vote accepted

$$\sqrt2^{\sqrt 3}<^?\sqrt3^{\sqrt 2}$$ 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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Yet 2<e<3, so how? –  Wok May 22 '12 at 9:03
    
It answers the question for $e^\pi$ or $\pi^e$. –  Wok May 22 '12 at 9:08
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@wok: try $\sqrt2$ and $\sqrt3$ (not $2$ and $3$). –  Did May 22 '12 at 11:54
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@Didier Ok, my mistake. –  Wok May 22 '12 at 12:17
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Hint: Use the Logarithm function.

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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. –  Sniper Clown May 22 '12 at 3:27
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You can add a similar idea to Robert's, if that is what you had in mind. –  Pedro Tamaroff May 22 '12 at 4:56
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This is a fine hint. The asker may have not thought at all about the logarithm function. –  Potato Jul 21 '12 at 5:44
    
You're just saying that because your name is Potato. –  The Chaz 2.0 Oct 18 '12 at 4:31
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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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$$\sqrt{2}^{\sqrt{3}} \approx 1.414^{1.732} \approx 1.822$$ $$\sqrt{3}^{\sqrt{2}} \approx 1.732^{1.414} \approx 2.174$$ $$\text{The rest is clear.}$$

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I assume the problem is meant to be solved without a calculator, which may not be easy from your solution. –  Joe May 23 '12 at 3:30
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hahah nice one! –  student May 23 '12 at 4:42
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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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