But if you happen to have n=2^c, then an algorithm with logarithmic complexity only needs c time. Thats why this is usually referred to as exponential speedup i

But if you happen to have n=2^c, then an algorithm with logarithmic complexity only needs c time. Thats why this is usually referred to as exponential speedup in complexity theory, just like from O(2^n) to O(n). More concretely if the first algorithm needs 1024 seconds, the second one will need only 10 seconds in both cases, so I think it makes sense.