Backlog

I plan to go into more detail concerning the power and log relation, since I guess my Logjam effort was too compact in this area.

So let’s start by examining the powers of a “base” \(b\) up to, and just beyond, 10:

We have chosen to go up to 10, because our aim is to get an estimate of the logarithms to base 10. If we were looking for logs to base 2, we would have cut off there instead. The length of the list depends on how close your base is to 1. The size of the base does not matter, except that the closer to 1, the more accurate our estimates will be. Up to a point — I have left the complete decimal of the calculated numbers, rather than trying to neaten it up with a fixed number decimal places showing to allow easy alignment on the decimal point. What we see in all its raggedy glory is the early onset of rounding (and conversion from binary) effects: \(1.1 \times 1.1 = 1.21\) and not \(1.2100000000000002\) as shown if you make the table with \(b = 1.1\)! So we need to be moderate in our expectations, and keep critical faculties on alert. A bit like living at the end of 2020, really.

Now to get logs to base 10, from our base \(b\) calculations we need:

\[\log_{10}x=\frac{\log_{b}x}{\log_{b}10}\]

Now the logs to base \(b\) are found by reversing the table, where our log is the “\(n\)” value for a given “\(x\)” (\(b^{n}\)):

If you choose \(b=1.1\) you will find \(\log_{b}10\) is somewhere between 24 and 25. Going to \(b=1.01\), you find after a lot of scrolling a \(\log_{b}10\) value between 231 and 232. Unless you are very careful, it is unlikely that your \(\log_{b}10\) is going to be a whole number (or even close to one). We could try fine-tuning the base, but then what about other values of \(\log_{b}x\)? Rather we assume/guess/estimate that \(\log_{b}10\) and \(\log_{b}x\) lie on the straight lines between adjoining points.

With \(b=1.1\), we have \(\left(x,\log_{b}x\right)\) values of \(\left(9.84\ldots,24\right)\) and \(\left(10.83\ldots,25\right)\). The straight line going through these points has the formula:

\[\log_{b}x\thickapprox24\frac{x-10.83\ldots}{9.84\ldots-10.83\ldots}+25\frac{x-9.84\ldots}{10.83\ldots-9.84\ldots}\]

This form should make it fairly easy to convince yourself in your head that when \(x=9.84\ldots\) we have \(\log_{b}x=24\), and the same for the other target value. The formula can be rearranged:

\[\log_{b}x\thickapprox\frac{25\left(x-9.84\ldots\right)-24\left(x-10.83\ldots\right)}{10.83\ldots-9.84\ldots}=\frac{x-25\left(9.84\ldots\right)+24\left(10.83\ldots\right)}{10.83\ldots-9.84\ldots}\]

Inserting \(x=10\), one finds \(\log_{b}x\thickapprox24.1525\ldots\). A more accurate computer calculated value is \(\log_{b}x\thickapprox24.1588\ldots\). Using the points closest to the integers up to 10, we use similar “linear interpolation” to derive an estimate to the table (and for other \(x\) values if desired):

These considerations raise the question: What if we allow the base to approach 1 more and more closely? The points that are interpolated get closer together, clearly a good thing (up to caveats due to binary conversion and rounding error). Another practical question would be the amount of computer time used as the number of multiplications to get up to 10 increases (assuming you were using some form of extra precision to avoid the caveats). These questions lead to the number e and natural logarithms, and need us to go into the head-bending ways of (mathematical) analysis. The couch has been prepared, please make yourself comfortable here . . .