I have read through questions like Calculate logarithms by hand and and a section of the Feynman Lecture series which talks about calculation of logarithms.
I have recognized neither of them useful for my purpose which is to quickly calculate logarithms of base $10$ upto $4$ digit accuracy
(I believe 4 is the goldilocks number in this case) .
I wish to find things like $\log_{10}(2) \approx 0.3010$ quickly without using a calculator or log table. Why? Because I want to be free from carrying them around and losing them all day. Plus, they're not always available when I need them (you can guess why).
My main purpose is to approximate the answers of very large and very small results of time consuming calculations.
Logarithms make that job much easier for me. For example,
$$\frac{87539319}{1729} \approx 10^{7.942 – 3.237} = 10^{4.705} = 5.069*10^4$$
According to Wolfram (Yup, I'm that lazy) the answer is, $50630.0\overline{283400809716599190}$.
Yes, I've over estimated by around $60$ but thanks to a log table, I did that approximation as fast as it took Wolfram to load the precise answer in my browser. But, without a log table, dividing itself would have me executing an iterative convergence just to find the multiples.
(1729*2 = too low, 1729*8= too high … this must be so intuitive for most of you)
So, a quick approximation method for logarithms would be really helpful to me.
Also, a good way to find antilogs will be nice as well.
I just realized that I can't compute decimal powers.
$$\Large 10^{0.3010} = 10^{0.3}*10^{0.001} = \sqrt[10]{1000} * \sqrt[1000]{10} = \text{Calculator Required}$$
I checked "
How to calculate a decimal power of a number"
but alas, the thing which came closest to what I needed required a calculator for an intermediate step. Defeats the purpose, I know.
If I can't find the antilog, the whole point of having a quick way to find the logarithm would be lost.
I hope you can help.
Best Answer
For quick-and-dirty approximations, I'm fond of the musical logarithms (writeup by Sanjoy Mahajan of work due to I. J. Good, who credits his father). Essentially this boils down to a mathematical fact:
and a "musical" fact:
I call this a "musical" fact because $2^{1/12}$ is the frequency ratio corresponding to an (equal-tempered) semitone.
For example: $3/2$ is the frequency ratio corresponding to the musical interval of a perfect fifth, which is seven semitones; thus $3/2 \approx 2^{7/12} \approx 10^{7/40}$ and so $\log_{10} 3/2 \approx 7/40$.