Question

If $x,y$ satisfy the equation $y^x=x^y$ and $x=2y$, then $x^2+y^2=r$, the value of $\frac{r}{4}$ is

Answer 1

$y^x=x^y$

If $x=2y$ then $y^{2y}=(2y{)}^y$

$\Rightarrow 2y\log y=y\log \left(2y\right)$

If $y\ne 0$ then $\log y^2=\log \left(2y\right)$

$\Rightarrow y^2=2y\Rightarrow y=2$

$\Rightarrow x^2+y^2=5y^2=20=r$

$\therefore \frac{r}{4}=\frac{20}{4}=5$