$x^x=y^y$ part 1

Let $y=rx$ where $r>1$ . 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 $\ln x=\frac{-r \ln r}{r-1}$ or equivalently, $x = e^{\frac{-r \ln r}{r-1}} = \left( e^{-\ln r}\right) ^{ {\frac{r }{r-1}}} = \left(\frac{1}{r}\right)^{ {\frac{r }{r-1}}}$ . With $x= \left(\frac{2}{7}\right)^{\frac{7}{5}}$ , we have $r=\frac{7}{2}$ .

Next, $y=rx = r \times \left(\frac{1}{r}\right)^{ {\frac{r }{r-1}}} = \left(\frac{1}{r}\right)^{ {\frac{1 }{r-1}}}=\left(\frac{2}{7}\right)^{ {\frac{2 }{5}}} = \left(\frac{4}{49}\right)^{ {\frac{1 }{5}}}\approx 0.60586$ .

Finally, $a+b = 4+49=\boxed{53}$ .