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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J. M. ain't a mathematician
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asked May 22, 2012 at 3:03
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1$\begingroup$ This again $e^\pi$ or $\pi^e$ $\endgroup$– checkmathMay 22, 2012 at 3:14
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2$\begingroup$ @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. $\endgroup$– RafidMay 22, 2012 at 5:33
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1$\begingroup$ @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! $\endgroup$– user641May 22, 2012 at 8:16
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6 Answers
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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 $\sqrt3^{\sqrt 2}$.
answered May 22, 2012 at 3:18
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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.
answered May 22, 2012 at 3:15
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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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13$\begingroup$ 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, 2012 at 3:27
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1$\begingroup$ You can add a similar idea to Robert's, if that is what you had in mind. $\endgroup$– Pedro ♦May 22, 2012 at 4:56
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1$\begingroup$ This is a fine hint. The asker may have not thought at all about the logarithm function. $\endgroup$– PotatoJul 21, 2012 at 5:44
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2$\begingroup$ You're just saying that because your name is Potato. $\endgroup$ 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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2$\begingroup$ I assume the problem is meant to be solved without a calculator, which may not be easy from your solution. $\endgroup$– JoeMay 23, 2012 at 3:30
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