Question

Решить систему: log(2)(x+y)+2log(4)(x-y)=5 3^1+2log(3)(x-y)=48 если не сложно , напишите, как можно подробней

Answer 1

Log₂(x+y)+2*log₄(x-y)=5    ОДЗ:   x>y    x>-y
3^(1+2*log₃(x-y)=48

log₂(x+y)+2*log₂²(x-y)=5
3*3^log₃(x-y)²=48

log₂(x+y)+2*(1/2)*log₂(x-y)=5
3*(x-y)²=48  |÷3

log₂(x+y)+log₂(x-y)=5
(x-y)²=16

1)
log₂((x+y)*(x-y))=5*log₂2
x-y=4

log₂(x²-y²)=log₂2⁵
y=x-4

x²-y²=32
y=x-4

x²-(x-4)²=32
x²-x²+8x-16=32
8x=48  |÷8
x=6   ⇒
y=6-4=2
2)
x²-y²=32
x-y=-4

x²-y²=32
y=x+4

x²-(x+4)²=32
x²-x²-8x-16=32
-8x=48  |÷(-8)
x=-6   ⇒
y=-2 ∉ ОДЗ
ответ: x=6    y=2.

Answer 2

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

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