Mr. Parthsharma ,the substitution you have done is not correct and complete.Let n = 3^{k}.So log_{3}n = k and n^{1/3} = 3^{k/3}.So the given recurrence becomes S(k) = S(k/3) + k which on solving gives Ø(k) = Ø(log_{3}n) .Hence Ø(log_{3}n) is the answer.So in your recurrence S(m) = S(m/3) + logm it should be m not logm.

T(n)=T(n^1/3) + logn

let n=2^m

T(2^m)=T(2^m/3) + log 2^m

again let T(2^m)=S(m)

then T(2^m/3) becomes S(m/3)

S(m)= S(m/3) + log m

we can easily solve this using masters method

Mr. Parthsharma ,the substitution you have done is not correct and complete.Let n = 3

^{k}.So log_{3}n = k and n^{1/3}= 3^{k/3}.So the given recurrence becomes S(k) = S(k/3) + k which on solving gives Ø(k) = Ø(log_{3}n) .Hence Ø(log_{3}n) is the answer.So in your recurrence S(m) = S(m/3) + logm it should be m not logm.see the solution of above question

T(n)=T(n

^{1/3})+lognn=2

^{m}where m=log_{2}nT(2

^{m})=T(2^{m/3})+mLet T(2

^{m})=S(m) and T(2^{m/3}) =S(M/3)so S(m)=S(m/3)+m

using master theoram 3rd case appled becase a<b

^{k}where a=1 ,b=3,k=1,p=0time complexity is O(m)

after subtitution the time complexity is O(logn)