Why is the log in the big-O of binary search not base 2?

Question

I am new to understanding computer science algorithms. I understand the process of the binary search, but I am having a slight misunderstanding with its efficiency.

In a size of $s = 2^n$ elements, it would take, on average, $n$ steps to find a particular element. Taking the base 2 logarithm of both sides yields $\log_2(s) = n$. So wouldn't the average number of steps for the binary search algorithm be $\log_2(s)$?

This Wikipedia article on the binary search algorithm says that the average performance is $O(\log n)$. Why is this so? Why isn't this number $\log_2(n)$?

Answer 1

When you change the base of logarithm the resulting expression differs only by a constant factor which, by definition of Big-O notation, implies that both functions belong to the same class with respect to their asymptotic behavior.

For example $$\log_{10}n = \frac{\log_{2}n}{\log_{2}10} = C \log_{2}{n}$$ where $C = \frac{1}{\log_{2}10}$.

So $\log_{10}n$ and $\log_{2}n$ differs by a constant $C$, and hence both are true: $$\log_{10}n \text{ is } O(\log_{2}n)$$ $$\log_{2}n \text{ is } O(\log_{10}n)$$ In general $\log_{a}{n}$ is $O(\log_{b}{n})$ for positive integers $a$ and $b$ greater than 1.

Another interesting fact with logarithmic functions is that while for constant $k>1$, $n^k$ is NOT $O(n)$, but $\log{n^k}$ is $O(\log{n})$ since $\log{n^k} = k\log{n}$ which differs from $\log{n}$ by only constant factor $k$.

Answer 2

In addition to fade2black's answer (which is completely correct), it's worth noting that the notation "$\log(n)$" is ambiguous. The base isn't actually specified, and the default base changes based on context. In pure mathematics, the base is almost always assumed to be $e$ (unless specified), while in certain engineering contexts it might be 10. In computer science, base 2 is so ubiquitous that $\log$ is frequently assumed to be base 2. That wikipedia article never says anything about the base.

But, as has already been shown, in this case it doesn't end up mattering.