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Algebra Difficulty 4.8 AIME Prove it Slovenia

Find all positive real numbers $x$ and $y$ such that
$x^{x+y} = y^{x-y} \quad \text{and} \quad x^2y = 1$

Solution

The second equation implies $y = x^{-2}$ , so
$x^{x + x^{-2}} = x^{-2(x - x^{-2})}.$
Taking the logarithm on both sides we get
$(x + x^{-2}) \log x = -2(x - x^{-2}) \log x.$
If $\log x = 0$ , then $x = 1$ and $y = 1$ . Otherwise,
$x + x^{-2} = -2x + 2x^{-2},$
so $3x^3 = 1$ . This implies $x = \frac{1}{\sqrt{3}}$ and $y = \sqrt{3}$ . There are two solutions: $x = 1$ , $y = 1$ and $x = \frac{1}{\sqrt{3}}$ , $y = \sqrt{3}$ .

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