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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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    @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!2012-05-22

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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    @Didier Ok, my mistake.2012-05-22
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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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    You're just saying that because your name is Potato.2012-10-18
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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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    hahah nice one!2012-05-23