According to this StackExchange answer https://cs.stackexchange.com/a/50356 the speedup is best expressed as a function relating the two algorithm's run times. So if there's a constant factor of 2, then that's "twice as fast", but if the relationship isn't linear (e.g. 2^(2n) = (2^n)^2 is a quadratic function of 2^n) then the speedup is named correspondingly. So 2^(2n) -> 2^n is quadratic speedup, while 2^n -> n or n -> log n are exponential speedups.