3. Find all positive real numbers $x$ and $y$ for which

$$
x^{x+y}=y^{x-y} \quad \text { and } \quad x^{2} y=1
$$
III/3. From the second equation, it follows that $y=x^{-2}$, so we have

$$
x^{x+x^{-2}}=x^{-2\left(x-x^{-} 2\right)}
$$

We take the logarithm of the equation and get

$$
\left(x+x^{-2}\right) \log x=-2\left(x-x^{-2}\right) \log x
$$

If $\log x=0$, then $x=1$ and thus $y=1$. Otherwise, we have

$$
x+x^{-2}=-2 x+2 x^{-2}
$$

or $3 x^{3}=1$. From this, it follows that $x=\frac{1}{\sqrt[3]{3}}$ and then $y=\sqrt[3]{9}$. We have obtained two solutions.

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-14.jpg?height=51&width=1633&top_left_y=1708&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-14.jpg?height=48&width=1633&top_left_y=1758&top_left_x=220)

Comparison of exponents, which is the equation $x+x^{-2}=-2 x+2 x^{-2}$ or the equivalent. 1 point Solution $x=\frac{1}{\sqrt[3]{3}}, y=\sqrt[3]{9}$