$x^x=y^y$ part 2

Let $y=rx$ where $r \neq 1$ . Now $x^x=y^y$ implies that $x \ln x = y \ln y = rx \ln(rx) = rx \left(\ln r + \ln x\right)$ . This means that $\ln x=\frac{r \ln r}{1-r}$ . Now $\displaystyle\lim_{r \to 1}\frac{r \ln r}{1-r} = \lim_{r \to 1}\frac{r \left(\frac{1}{r}\right) + \ln r}{-1} =-1$ and so $p=\displaystyle\lim_{r \to 1} x = e^{-1} \approx 0.36787944$ .

Hence $\lfloor 100000p \rfloor = \boxed{36787}$ .

$x^x=y^y$

$x ln x = y ln y$

Let $f(x) = x ln x$

By evaluating graph of f(x), x and y can take different values to each other if a horizontal line cuts the graph twice. We seek a value where x=y and a horizontal line is tangent to the graph.

So: $f'(x)=1 + ln x$ which gives $f'(x)=0$ for $x = \frac{1}{e}$

$p=\frac{1}{e}$