Решить систему: log(2)(x+y)+2log(4)(x-y)=5 3^1+2log(3)(x-y)=48 Если не сложно , напишите,...

$\displaystyle \left \{ {{log_2(x+y)+2log_4(x-y)=5} \atop {3^{1+2log_3(x-y)}=48}} \right.\\\\log_2(x+y)+2* \frac{1}{2} log_2(x-y)=5\\\\log_2(x+y)(x-y)=5\\\\3*3^{log_3(x-y)^2}=48\\\\(x-y)^2=16$
$\displaystyle \left \{ {{log_2(x+y)(x-y)=5} \atop {(x-y)^2=16}} \right.$

1.
$\displaystyle \left \{ {{(x-y)(x+y)=2^5} \atop {x-y=4}} \right.\\\\ \left \{ {{4(x+y)=32} \atop {x-y=4}} \right.\\\\ \left \{ {{x+y=8} \atop {x-y=4}} \right.\\\\ \left \{ {{x=6} \atop {y=2}} \right.$

2.
$\displaystyle \left \{ {{(x-y)(x+y)=32} \atop {x-y=-4}} \right.\\\\ \left \{ {{-4(x+y)=32} \atop {x-y=-4}} \right.\\\\ \left \{ {{x=-6} \atop {y=-2}} \right.$

решение не подходит под ОДЗ

ответ х=6, у=2