Here we have to analyse like this :
For ith value , the inner loop runs log(i) times.
So say when i = 2 , the inner loop runs log(2) times , when i = 3 , the inner loop runs log(3) times and so on till i = n..
So time complexity = log(1) + log(2) + log(3) ........ + log(n)
= log(1.2.3.....n)
= log(n!)
= nlogn - n + O(logn) = nlogn [ By Stirling's approximation and considering the dominant term ]
= θ(nlogn)
Hence time complexity = θ(nlogn)