I understand the idea intuitively that because the number of options for each digit is more "granular" for base $2$, we need more digits to represent the same number as opposed to base $10$.
How can I express this idea (or another proof) in mathematical terms?
Take an integer $n$, for some $p\in \mathbb{Z}$ it is true that $2^p < n\leq 2^{p+1}$. It is clear that $p+1=\lceil \log_{2}(n)\rceil$. Similarly, for some $q\in \mathbb{Z}$ it is true that $10^q< n\leq 10^{q+1}$ and therefore $q+1=\lceil \log_{10}(n)\rceil$. You can see that the first quantity is always at least as large as the second quantity which is what we want to show (The first quantity is the length of the binary representation and the second the length of the decimal representation).