Master theorem: t(n) = 3t(n/3) + o(log n)? hello, i've been trying to solve the bounds for the recurrence stated in the title: i know a = 3, b = 3, and f(n) = o(log n), n^log(a) of base b --> n^log(3) of base 3 --> n^1 = n. so i get θ(n). now i proceed to use case 1 of master theorem: if f(n) = o(n^log(a-e) of base b) for some constant e > 0, then t(n) = θ(n^log(a) of base b) since f(n) < n^log(a) of base b, asymptotically. i believe the solution should be θ(n) but where i am stuck is proving that f(n) is polynomially smaller, when using case 1, the ratio of f(n) / n^log(a) of base b: = log n / n, so that the solution θ(n) is true any with this is appreciated