For T(n)= aT(n/b)+n^k log^p n
for a>=1, b>1, k>=0,p real no.
compare a and b^k,
1. if a>b^k then T(n)=n^(log a/log b)
2. if a=b^k then
.if p>-1, T(n)=n^(log a/log b) log^(p+1) n
.if p=-1, T(n)=n^(log a/log b) log log n
.if p<-1, T(n)=n^(log a/log b)
3. if a<b^k then
.if p>=0, T(n)=n^k log^p n
.if p<0, T(n)=n^k
compare a and b^k i.e a=2, b^k=2^1 which implies a=b^k
Since, p >=(-1), so, T(n)=n log^2 n