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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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Thanks for you advice. – Vinicius L. Beserra Sep 29 '12 at 19:14

2 Answers 2

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

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

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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
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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