Nolinear system of equations.

Question

$$ \left\{\begin{array}{l} x^y = y^x \\ x-y\cdot\log_xy=(x+y)\cdot\log_xy \end{array} \right. $$

Thanks for your time!

Answer 1

With rules for logarithms the second equation translates to $$ x - y\log_xy = \log_x y^x + y \log_x y $$ and using the first equation this simplifies to $$ x= y+2y (1+\log_xy) $$ and then $$ \log_xy = \frac{x-y}{2y} $$ From first equation you can express $$ y = x^{\frac{y}{x}} $$ implying $$ \frac{y}{x} = \frac{x-y}{2y} $$ which translates to $$ x^2 -xy - 2y^2 = 0 $$ and factoring this you get $$ (x+y)(x-2y)=0 $$ that gives $2$ solutions which, as usual with equation with logarithms, need to be checked. Try.