To address this question, we utilize a key property of logarithms. Specifically, we employ the property that states:
[tex]\[
\log(a) - \log(b) = \log\left(\frac{a}{b}\right)
\][/tex]
Given Benford's law formula:
[tex]\[
P(d) = \log(d+1) - \log(d)
\][/tex]
We can rewrite the expression for [tex]\( P(d) \)[/tex] as a single logarithm by applying the logarithm property mentioned above. Therefore, we have:
[tex]\[
P(d) = \log\left(\frac{d+1}{d}\right)
\][/tex]
This simplification now allows us to find the probability that the number 1 is the leading digit. We substitute [tex]\( d = 1 \)[/tex] into the simplified expression:
[tex]\[
P(1) = \log\left(\frac{1+1}{1}\right) = \log\left(\frac{2}{1}\right) = \log(2)
\][/tex]
To determine [tex]\( P(1) \)[/tex], we need the value of [tex]\(\log(2)\)[/tex], which is approximately:
[tex]\[
\log(2) = 0.6931471805599453
\][/tex]
Thus, the probability that the number 1 is the leading digit according to Benford's law is approximately [tex]\( 0.6931471805599453 \)[/tex].
