2^x*2^y=16 log3x+log3y=1 решите пожалуйста систему уравнений с:

[latex] \left \{ {{2^x*2^y=16} \atop { log_{3}x+ log_{3}y=1 }} \right. [/latex] ОДЗ: [latex]x\ \textgreater \ 0,[/latex]   [latex]y\ \textgreater \ 0[/latex] [latex] \left \{ {{ 2^{x+y} =2^4} \atop { log_{3}(xy)= log_{3}3 }} \right. [/latex] [latex] \left \{ {{ {x+y} =4} \atop { xy=3 }} \right. [/latex] [latex] \left \{ {{ {x} =4-y} \atop { y(4-y)=3 }} \right. [/latex] [latex] \left \{ {{ {x} =4-y} \atop {4y-y^2-3=0 }} \right. [/latex] [latex] \left \{ {{ {x} =4-y} \atop {y^2-4y+3=0 }} \right. [/latex] [latex]y^2-4y+3=0[/latex] [latex]D=(-4)^2-4*1*3=16-12=4[/latex] [latex]y_1= \frac{4+2}{2}=3 [/latex],     [latex]x_1=4-3=1[/latex] [latex]y_2= \frac{4-2}{2}=1 [/latex],      [latex]x_2=4-1=3[/latex] Ответ: (1;3); (3;1)