log(1/5)*(3*x - 2) < log(1/5)*(6 - x)
(3*x - 2)*log(1/5) < (6 - x)*log(1/5)
(2, oo)
x in Interval.open(2, oo)
And(2 < x, x < oo)
(2 < x)∧(x < oo)