Question

2log²(по основанию 3)(x-1) + 3log(по осн 3)(x-1) + 1 + log(по осн 3)2 <= log(по осн 3)54

Answer 1

ОДЗ
x-1>0⇒x>1
log(3)(x-1)=a
2a²+3a+1≤log(3)54-log(3)2
2a²+3a+1≤log(3)(54/2)
2a²+3a+1≤log(3)27
2a²+3a+1≤3
2a²+3a+1-3≤0
2a²+3a-2≤0
D=9+16=25
a1=(-3-5)/4=-2
a2=(-3+5)/4=1/2
-2≤a≤1/2
-2≤log(3)(x-1)≤1/2
1/9≤x-1≤√3
1 1/9≤x≤√3+1
x∈[1 1/9;√3+1]