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Trivial nitpick: since this is binary search, the base for the logarithm is 2, not e. So it would be 2^M.

Nitpick upon nitpick, the base may be 2 rather than e but IIRC the difference between lg(n) and log2(n) is a constant, and the log/lg term dominates, so it still is 0(log). I think.

Big fleas have little fleas upon their backs to bite 'em \ And little fleas have lesser fleas, and so, ad infinitum.

(https://en.wikipedia.org/wiki/The_Siphonaptera>

Nitpick on Nitpick on Nitpick, it is indeed O(2^m) and not O(e^m). Those are not the same complexity.

Logs get to not care, but when we exponentiate it matters.

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