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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 $y^x=\frac{1}{\sqrt x}$

2 Answers 2

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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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    Oh sorry I overlooked.2012-09-30