Logarithms and the history from Babbage's time are truly fascinating.
So we start with the Binomial Expansion noting this is only useful between 0 and 1:
$$ \ln x = (x-1) - \frac{1}{2}(x-1)^2 + ....$$
We'll do this for a sixth order polynomial:
$$ \ln x = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 + \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 + \frac{1}{6}(x-1)^6 $$
As an excel function - this looks like this:
=(C161-1)-(1/2)*POWER((C161-1),2)+(1/3)*POWER((C161-1),3)-(1/4)*POWER((C161-1),4)+(1/5)*POWER((C161-1),5)-(1/6)*POWER((C161-1),6)
To understand sum of the differences we look at the following example:
\begin{array}{|c|c|c|c|}
\hline
x& p(x) = 2x^2 − 3x + 2 & diff1(x) = ( p(x + 1) − p(x) ) & diff2(x) = ( diff1(x + 1) − diff1(x) ) \\ \hline
0 & 2& -1&4\\ \hline
1 & 1 & 3 &4\\ \hline
2 & 4 & 7&4\\ \hline
3 & 11 & 11 &\\ \hline
4 & 22 & &\\ \hline
\end{array}
ie you keep adding the number in the right column to the next column over, and each row you carry the sum over to the right. (You can see a detailed video here).
Now we build a table for this polynomial of sum of the differences between 0 and 1:
\begin{array}{|c|c|c|c|c|c|c|}
\hline
index & binomial calculation & 1st diff & 2nd diff & 3rd diff & 4th diff & 5th diff & 6th diff \\ \hline
1&0.00000&&&&&&\\ \hline
0.9&-0.10536&-0.10536&&&&&\\ \hline
0.8&-0.22314&-0.11778&-0.01242&&&&\\ \hline
0.7&-0.35663&-0.13349&-0.01571&-0.00329&&&\\ \hline
0.6&-0.51046&-0.15383&-0.02034&-0.00463&-0.00134&&\\ \hline
0.5&-0.69115&-0.18068&-0.02685&-0.00651&-0.00188&-0.00054&\\ \hline
0.4&-0.90773&-0.21658&-0.03590&-0.00905&-0.00254&-0.00066&-0.00012\\ \hline
0.3&-1.17258&-0.26485&-0.04827&-0.01237&-0.00332&-0.00078&-0.00012\\ \hline
0.2&-1.50229&-0.32971&-0.06486&-0.01659&-0.00422&-0.00090&-0.00012\\ \hline
0.1&-1.91870&-0.41640&-0.08669&-0.02183&-0.00524&-0.00102&-0.00012\\ \hline
\end{array}
Now we shuffle top of columns to top row
\begin{array}{|c|c|c|c|c|c|c|}
\hline
index & binomial calculation & 1st diff & 2nd diff & 3rd diff & 4th diff & 5th diff & 6th diff \\ \hline
1&0.00000&-0.10536&-0.01242&-0.00329&-0.00134&-0.00054&-0.00012\\ \hline
\end{array}
Then we fill in first row with precomputed values, then fill down by adding the number above and above right
\begin{array}{|c|c|c|c|c|c|c|}
\hline
index & binomial calculation & 1st diff & 2nd diff & 3rd diff & 4th diff & 5th diff & 6th diff \\ \hline
1&0.00000&-0.10536&-0.01242&-0.00329&-0.00134&-0.00054&-0.00012\\ \hline
0.9&-0.10536&-0.11778&-0.01571&-0.00463&-0.00188&-0.00066&-0.00012\\ \hline
0.8&-0.22314&-0.13349&-0.02034&-0.00651&-0.00254&-0.00078&-0.00012\\ \hline
0.7&-0.35663&-0.15383&-0.02685&-0.00905&-0.00332&-0.00090&-0.00012\\ \hline
0.6&-0.51046&-0.18068&-0.03590&-0.01237&-0.00422&-0.00102&-0.00012\\ \hline
0.5&-0.69115&-0.21658&-0.04827&-0.01659&-0.00524&-0.00114&-0.00012\\ \hline
0.4&-0.90773&-0.26485&-0.06486&-0.02183&-0.00638&-0.00126&-0.00012\\ \hline
0.3&-1.17258&-0.32971&-0.08669&-0.02821&-0.00764&-0.00138&-0.00012\\ \hline
0.2&-1.50229&-0.41640&-0.11490&-0.03585&-0.00902&-0.00150&-0.00012\\ \hline
0.1&-1.91870&-0.53130&-0.15075&-0.04487&-0.01052&-0.00162&-0.00012\\ \hline
\end{array}
Now we have a value for ln 0.3 = -1.1725805
Now to convert ln 0.3 to ln 3 we add ln 10 which is approx 2.302
And we get ln 3 = 1.1294195 which is close to a computed value of ln 3 = 1.098612289
