If you pick base 2 for the then log2(10000) is about 13. So if you are a constant factor of >13 faster for the work that dominates, the n*log(n) will be faster

If you pick base 2 for the then log2(10000) is about 13. So if you are a constant factor of >13 faster for the work that dominates, the n*log(n) will be faster. Slightly OT: I remember in my algorithms class we used to joke that log(log(N)) was the same as a constant factor of 5 (in the 64-bit era, perhaps we should say 6).