# Please help me to remove Feynman's $\log_{10}$ from my eye.

I am enjoying the Feynman Lectures, Volume I, Chapter 22, particularly section 22-4, wherein Feynman generates the natural logarithm base from bamboo and coconuts. I must confess, I got a little lost on Table 22-1, where the narrative seems a little hand-wavy with respect to the column, $(\mathrm{10}^{s}-1)/s$.

I see that this takes the bit after the decimal point in the power of ten and divides it by the power. Perhaps that would give us the rate of increase of bit after the decimal, the slope on the "exponent vs. the-bit-after-the-decimal" graph as the exponent approaches zero, how much that bit increases per increase in exponent?

I'm definitely missing the point. Cans someone shed some light on this?

I fail to follow Feyman's heuristics (and I'm not sure if there is something there).

When he says that there is a better method, what he is doing (without telling you) is to write, for $\Delta$ small, $$\tag{1} 10^\Delta=e^{\Delta\,\log 10}\simeq1+\Delta\log 10.$$ So he says that this is a better approximation: $\log10$ is the number "$2.3025$" that he mentions.

The fourth column is a method to find $\log 10$: you have $$\frac{10^s-1}s=\frac{e^{s\log 10}-1}s=\frac{1+s\log 10+o(s^2)-1}s=\log10+o(s).$$ Thus the numbers in the fourth column indeed converge to $\log10$, which can be used for the approximation $(1)$.

I could not fully understand your question, but let me explain what I understand from the table and work of Feynman. Hope it helps.

Step 1) Calculate $10^s$ column. This is done by taking square roots.

Step 2) Make important observation:
Instead of continuing step 1, observe that $$\frac{10^s - 1}{s} \to c$$ goes to some constant c. And crucial observation is changes are $211,104,53,26$; all of them are approximately half of the previous one. Therefore, without any further calculation, it will change approximately 26. Here I believe it might be useful to note that; $$2.3234 - 0.0211 = 2.3023\\ 2.3130 - 0.0104 = 2.3026\\ 2.3077 - 0.053\ \ = 2.3024\\ 2.3051 - 0.026\ \ = 2.3025$$ where the last one is used.

Step 3) Since we determine $c$ as $2.3025$, solve it back. $$10^s = 1+2.3025s$$ and let $s = \Delta/1024$; $$10^s = 1+0.0022486\Delta$$

As it continues; to calculate logarithm of $2$, he notes that 2 can be written as product of numbers in column 3, where we already calculated them by taking square roots and a last bit is too small and not in column 3. There comes the role of $\Delta$. He solves; $$1.000573 = 1 + 0.0022486\Delta \implies \Delta = 0.254$$ Note that this correction is of order 4 figures, and ables him to approximate $s$ up to 5 figures.