Solve the following system of equations ($x,y \in \Bbb R$):
$$\begin{cases} 3^{x+3y-2} + 6\cdot 3^{y^2+4x-2} &=5^{5y-3x} + 2\cdot 3^{y^2-2y+1}\\ 1+2\sqrt{x+y-1} &=3\sqrt[3]{3y-2x}. \end{cases}$$
I think about it but I still have no solution... :(
Since the second equation, I write $x+y \ge 1$ then $y \ge \dfrac{2}{3}x.$ So $y \ge \dfrac{2}{5}$.
I rewrite the 1st equation: \begin{align*} 3^{x+3y-2} + 6\cdot 3^{y^2+4x-2} &=5^{5y-3x} + 2\cdot 3^{y^2-2y+1}\\ \iff 3^{x+3y-2} + 6\cdot 3^{y^2+4x-2}& \le 5^{5y-3x} -3^{5y-3x}+3^{5y-3x} + 2\cdot 3^{y^2+2y+1}\\ \iff (3^{x+3y-2}-3^{5y-3x})(1+2\cdot 3^{y^2+3x-3y+1})& \le 5^{5y-3x} -3^{5y-3x}\\ \iff (9^{2x-y-1}-1)\underset{>0}{\underbrace{(1+2\cdot 3^{y^2+3x-3y+1}})}&\le \left (\dfrac{5}{3} \right )^{5y-3x}-1. \end{align*} Now, I have trouble.... Can anyone post the roots of this system of exponential equations.
I really appreciate if some one can help me. Thanks!