1)вычислить 1/3 log2/3 8-3log2/3 3+log2/3 6 2) решить уравнение log 1/7(3x+4)=-1

1
1/3*log(2/3)8-3log(2/3)3+log(2/3)6=log(2/3)∛8-log(2/3)3³+log(2/3)6=
=log(2/3)2-log(2/3)27+log(2/3)6=log(2/3)(2*6/27)=log(2/3)(4/9)=2
2
log(1/7)(3x+4)=-1
{3x+4>0⇒x>-4/3
{3x+4=7⇒3x=3⇒x=1
ответ х=1