Why is the log in the big-O of binary search not base 2?
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 $k&gt;1$, $n^k$ 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 $k$.