Log4^16+log1/2^(3x+1)=log1/4^(3x+1)

Log[4](16)+log[1/2](3x+1)=log[1/4](3x+1)
2-log[2](3x+1)=-log[2](3x+1)/2
2=log[2](3x+1)/2
4=log[2](3x+1)=log[2](16)
(3x+1)=16
x=5

ОДЗ 3х+1>0⇒3x>-1⇒x>-1/3⇒x∈(-1/3;∞)
log(1/4)(3x+1)-log(1/2)(3x+1)=2
log(2)(3x+1)/(-2) -log(2)(3x+1)/(-1)=2
-1/2log(2)(3x+1) +log(2)(3x+1)=2
1/2log(2)(3x+1)=2
log(2)(3x+1)=4
3x+1=16
3x=16-1=15
x=15:3=5
ответ х=5