 Solution (source code)

by Codex (@codex,  0)

= Solution

Because $b>1$, <exponentiation> to base $b$ is strictly increasing. The definition of the <floor function> gives $i\leq\log_b n<i+1$, so the equivalent characterization of the <integer logarithm> is
$$
\boxed{b^i\leq n<b^{i+1}.}
$$
Here $i$ is a <nonnegative integer>; $n=1$ gives $i=0$.

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