Incidentally, while it focuses on values that fit in machine registers, you might check out Hacker's Delight's section on computing "ilog10". In the second edi

Incidentally, while it focuses on values that fit in machine registers, you might check out Hacker's Delight's section on computing "ilog10". In the second edition, section 11-4 is on integer logarithms. If your target language is Python3, you have the "bit_length()" method of integer objects. Multiply by c ~= 0.30103 and then correct for being too low; superficially, it seems like this algorithm should work for extremely large numbers, at least to millions of digits. As far as test cases, n10, (n10)+1 and (n10)-1 seem like good choices and as long as the underlying mathematical function is monotonic should ferret out failures. I could easily run my implementation for numbers up to 30,000 digits, but going to 300,000 digits didn't finish while writing this message. log10_2 = math.log10(2) def digit_length_clever3(n): if not n: return 1 est = math.floor(n.bit_length() * log10_2) low = n >= 10**est return est + low