C:\Users\Peter>perl -MPOSIX=log10 -le "$,=', ';print $_, int($_) for l
+og(1e12)/log(10), log10(1e12)"
12, 11
12, 12
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log() is implemented as precisely as possible for base e; unfortunately, dividing one log-base-e by another is not always as precise as you want; OTOH, using POSIX::log10, you get the function implemented as precisely as possible for base 10, which appears to be what you want.