Question

2^х*2^у=16
log3х (три снизу)+log3у (три снизу)=1
Решить систему уравнений

Answer 1

2^x*2^y=16         ОДЗ: x>0  y>0     2^(x+y)=2^4    x+y=4      y=4-x
log₃x+log₃y=1                log₃(x*y)=log₃3     x*y=3   x(4-x)=3   x²-4x+3=0  D=4
x₁=3    x₂=1
y₁=1    y₂=3.

Answer 2

$\left \{ {{ 2^{x }* 2^{y} =16 } \atop { log_{3}x+log_{3}y=1 }} \right. \\ \\ \left \{ {{ 2^{x+y}= 2^{4} } \atop {log_{3}(xy)=log_{3}3}} \right. \\ \\ \left \{ {{xy=3} \atop {x+y=4}} \right. \\ \\ \left \{ {{yx=3} \atop {x=4-y}} \right. \\ \\ \left \{ {{y(4-y)=3} \atop {x=4-y}} \right. \\ \\ \left \{ {{x=4-y} \atop {- y^{2}+4y-3=0 }} \right. \\ \\ \left \{ {{ y^{2}-4y+3=0 } \atop {x=4-y}} \right.$

$1. \left \{ {{y=3} \atop {x=4-3}} \right. \\ \\ \left \{ {{y=3} \atop {x=1}} \right. \\ \\ (1;3)$

$2. \left \{ {{y=1} \atop {x=4-1}} \right. \\ \\ \left \{ {{y=1} \atop {x=3}} \right. \\ \\ (3;1)$

Ответ: (3;1);(1;3)