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