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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
    May 22 '12 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
    May 22 '12 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
    May 22 '12 at 8:16
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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
    May 22 '12 at 9:03
  • $\begingroup$ It answers the question for $e^\pi$ or $\pi^e$. $\endgroup$
    – Wok
    May 22 '12 at 9:08
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    $\begingroup$ @wok: try $\sqrt2$ and $\sqrt3$ (not $2$ and $3$). $\endgroup$
    – Did
    May 22 '12 at 11:54
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    $\begingroup$ @Didier Ok, my mistake. $\endgroup$
    – Wok
    May 22 '12 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. $\endgroup$ May 22 '12 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 Tamaroff
    May 22 '12 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
    Jul 21 '12 at 5:44
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    $\begingroup$ You're just saying that because your name is Potato. $\endgroup$ Oct 18 '12 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
    May 23 '12 at 3:30
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    $\begingroup$ hahah nice one! $\endgroup$
    – lesnikow
    May 23 '12 at 4:42

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