[rand_r]

One part is that Big-O is an upper bound that kicks in for sufficiently large n.

For example, a O(log(n)) function will be faster than a O(k * n) function, for any fixed k, when n>j (for some fixed j i.e. when n is large enough). When comparing major classes of functions (n, log(n), 2^n, etc,), constant multiples don't matter, mathematically.

The other reason is that when comparing two functions with a similar amount of constant steps, let's say 5n vs 2n, it's impossible to say which is faster in the real world, so it's simpler and just as useful to collapse the set of O(k * f(n)) functions into O(f(n)).