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I have seen this question before, but I just can solve finding $x=y=1$, but the book tells me another answer, where $x=\frac{1}{2}$ and $y=2$.

This is the question. Find the solution of the system of equations where $x^y=\frac{1}{y^2}$ and $y^x=\frac{1}{\sqrt x}$

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The fact that you have a 0% acceptance rate is very bad. This means that you have never accepted any answers. When people take their time to answer your questions they can expect feedback. It is your duty to supply that feedback (and even some thanks!) by choosing a "best answer". If I were you then I'd go into your profile and go through your old questions and choose best answers. Your profile shows seven questions that you've posted. –  Fly by Night Sep 29 '12 at 18:43
    
Thanks for you advice. –  Vinicius L. Beserra Sep 29 '12 at 19:14
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2 Answers

up vote 6 down vote accepted

Rewrite these as $x^y = y^{-2}$ and $y^x = x^{-1/2}$, then express $x$, say: $$x=(y^x)^{-2}$$ then plug it into the other: $$(y^x)^{-2y} = y^{-2} $$ then either $y=1$ or $-2xy=-2$ and you can do similarly for the other one, with base $x$.

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$$I\;\;\;x^y=y^{-2}$$

$$II\;\;\;y^x=x^{-1/2}$$

$$ x^{yx}=y^{-2x}=(y^x)^{-2}=(x^{-1/2})^{(-2)}=x\Longrightarrow xy=1\Longrightarrow$$

$$I\;\;\;\Longrightarrow x^{1/x}=(x^{-1})^{-2}=x^2\Longrightarrow x=\frac{1}{2}\,\,,\,\,y=2$$

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That last step is incomplete. $x$ can be 1. –  mick Sep 30 '12 at 14:30
    
Yes, of course: the trivial answer $\,x=y=1\,$ that the OP already found before. –  DonAntonio Sep 30 '12 at 16:49
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Oh sorry I overlooked. –  mick Sep 30 '12 at 16:56
    
Thanks for all responses. –  Vinicius L. Beserra Jan 3 '13 at 17:19
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