Find the positive solutions of the system of equations $x^{x + y} = y^{n}$ and $y^{x + y} = x^{2 n} y^{n}$ where $n > 0$ .

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Solution

$x^{x + y} = y^{n}$ $(1)$
$y^{x + y} = x^{2 n} y^{n}$ . $(2)$
Multiplying the above, we get
$(x y)^{x + y} = (x y)^{2 n}$
$∴ x + y = 2 n$ .. $(3)$
Hence from $(1)$ , $x^{2 n} = y^{n}$ or $(x^{2})^{n} = y^{n}$ $∴ x^{2} = y$
Putting for y in $(3)$ , we get
$x^{2} + x - 2 n = 0$
$∴ x = \frac{- 1 \pm \sqrt{1 + 8 n}}{2}$
Since we have to find positive solution, x must be $+$ ve.
$∴ x = \frac{- 1 + \sqrt{1 + 8 n}}{2}$
$∴ y = 2 n - x = 2 n - \frac{- 1 + \sqrt{1 + 8 n}}{2}$
$= \frac{4 n + 1 - \sqrt{1 + 8 n}}{2}$ .

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