Benford's law, first digit at most 3

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With Benford’s law (probability that the first significant digit is $d$ equal to $\log_{10}(1+1/d)$), what is the probability that the first digit is $1$, $2$ or $3$ (i.e. $\le 3$)?

Solution

Add the probabilities of the three digits, using the properties of logarithms: $P=\log_{10}2+\log_{10}\tfrac32+\log_{10}\tfrac43=\log_{10}\!\left(2\cdot\tfrac32\cdot\tfrac43\right)=\log_{10}4\approx \boxed{0.602}.$ The first three digits alone therefore cover over $60\%$ of the cases.