Question

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

Answer 1

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

ответ: (1;3); (3;1)