Computing Base Ten Logs without Tables

Artemas Martin *St Andrews Un.

I've been researching some self-taught American mathematicians from the 19th Century, including Artemas Martin who edited (and typed and type-set and did all the mailing and collections) for  'The Mathematical Visitor' (1878 - 1894) and 'The Mathematical Magazine' (1882 - 1884).  
He eventually was recognized by mathematical organizations around the world, and received honorary degrees from Yale, Rutgers, and Hillsdale College in Michigan.
 One of the short pamphlets he wrote for the International Congress of Mathematics in Paris in 1900 described a method of finding base ten logarithms by hand.
It seemed to be so simple that I thought I would share it now that everyone has a calculator that will do it in no time.  But still, you may, like me, want to know a way by hand for those napkin conversations at the bar when you don't want to pull out your i-phone and talk about your new calculator app...
I have shortened the method so my results are more approximations than used by Mr Martin, but good approximations in most cases.  Check them out yourself.

The method used two rather well known infinite series for the base ten logarithm:

\(log_{10}{10^m + x} =  m + K(\frac{x}{10^m} - \frac{1}{2}(\frac{x}{10^m})^2\))

and \(log_{10}{10^m - x} =  m - K(\frac{x}{10^m} + \frac{1}{2}(\frac{x}{10^m})^2\))

The constant is the modulus of the common logs, appx .4343.
So if we wanted the log of 108, which is \(10^2 + 8\)  we would use
   \(log_{10}{10^2 + 8} =  2 + (.4343(\frac{8}{10^2} + \frac{1}{2}(\frac{8}{10^2})^2)\)

which simplifies to  \(=  2 + (.4343(.08 - .0032)) = 2+.4343(.0768) \)
Which I get (multiplying with pen and paper) to be 2.03335424 
My calculator gives \( log_{10}{1008}= 2.033423...\) So we are good to three decimal places

Greater accuracy could be obtained by adding the third term, \( \frac{1}{3} (\frac{x}{10^m})^3 \) to both the above series.

For smaller numbers it is necessary to find a power that will get us close to a value of 10^m

For example, to find \(\ log_{10}{5}\) we might use \( 5^6 = 15625 \)  and divide the answer by six... and so . rounding 5625 to 5000 (hey, we're approximating here)...

\(log_{10}{10^4 + 5000} =  4 + K(\frac{5000}{10^4} - \frac{1}{2}(\frac{5000}{10^4})^2\))
Conveniently \(\frac{5000}{10^4} = \frac{1}{2}\)  making our work somewhat easier  and we get
\(log_{10}{10^4 + 5000} =  4 + K(\frac{1}{2} - \frac{1}{2}(\frac{1}{2})^2\))
which leads us to\( 4 + .4343 (\frac{3}{8})\)    or 4.08604,,, And this is for 5⁶ so we divide by six to get the base ten log of 5, which should be about .681006   , and even with all the rounding we did (like using 5000 instead of 5600 or 5625 for calculating efficiency, we still have a relatively close answer. (maybe 6000 would have been a slightly better appx.?) 

I'm not ready to give up my calculator just yet, but if you walk up to me on the street when I don't have one (I am so waiting for this kind of thing to happen>>>) and ask me for the base ten log of 12, or 103, or 1005.... watch my smoke.....

Mr. Martin's paper is free on the net here, so read the original and try it for yourself.