Question

1.3log3^2x+log3x-4=0 помогите плиз
2.{log13(x^2+y^2)=2
log5x=log5y+1-log5^12

Answer 1

1
x>0
log(3)x=a
3a²+a-4=0
D=1+48=49
a1=(-1-7)/6=-4/3⇒log(3)x=-4/3⇒x=1/3∛x
a2=(-1+7)/6=1⇒log(3)x=1⇒x=3
2
{log(13)(x²+y²)=2⇒x²+y²=169
{log(5)x-log(5)y=log(5)5-log(5)12⇒log(5)(x/y)=log(5)(5/12)⇒x/y=5/12
12x=5y
y=2,4x
x²+(2,4x)²=169
x²+5,76x²=169
6,76x²=169
x2=169/6,76
x=-13/2,6=-5⇒y=2,4*(-5)=-12
x=13/2,6=5⇒y=2,4*5=12
(-5;-12);(5;12)