In this article, all logarithms and exponents are to base 10, and decimal answers are rounded appropriately.
The logarithm of a number is the power to which 10 must be raised to equal that number. Some simple examples:
- \(10^2 = 100\), therefore \(\log 100 = 2\)
- \(10^3 = 1000\), therefore \(\log 1000 = 3\)
- \(\log 200 = 2.301\) (between \(\log 100\) and \(\log 1000\))
Logarithms are used often in Mathematics, but rarely in mental calculation—partly because of their difficulty! In fact, the most advanced method I’ve taught when giving various classes on mental math is to use logarithms (calculated mentally) to calculate difficult roots and powers.
Mathematics:
Mathematicians learn that the following are always true:
- \(\log (a*b) = \log a + \log b\)
- \(\log (\frac{a}{b}) = \log a – \log b\)
Therefore if we know the approximate values \(\log 2 = 0.30103\) and \(\log 3 = 0.47712\) we can calculate the following examples:
- \(\log{20} = \log{10} + \log{2} = 1 + 0.30103 = 1.30103\)
- \(\log{6} = \log{2} + \log{3} = 0.30103 + 0.47712 = 0.77815\)
- \(\log{32} = 5 * \log{2} = 5 * 0.30103 = 1.50515\)
- \(\log{2.5} = \log{10} – 2 * \log{2} = 1 – 2 * 0.30103 = 0.39794\)
Our method for calculating logarithms depends on memorizing a variety of logarithm values, and combining these to find the value of the required logarithm.
Preparation:
Learn the following logarithm values – and optionally more:
\(\log 2 = 0.30103\)
\(\log 3 = 0.47712\)
\(\log 7 = 0.84510\)
\(\log 1.1 = 0.04139\)
\(\log 1.3 = 0.11394\)
\(\log 1.7 = 0.23045\)
\((\log 1.01 = 0.00432)\)
Method:
- Choose a number very close to the original number that you can construct using factors whose logarithms you have learnt.
- Calculate the logarithm of this approximated number
- Calculate the percentage error and multiply by \(\log{1.01} = 0.00432\) and add or subtract as appropriate
Examples:
Straightforward example with no approximation: \(\log{510}\)
This is a straightforward example as \(510 = 1.7 * 3 * 100\) and no approximation is needed.
\(\log 510 = \log{1.7} + \log{3} + \log{100} = 0.23045 + 0.47712 + 2 = 2.70757\)
Example that requires approximation: \(\log{511.28}\)
There are several sensible approaches here:
\(511.28 = 510 + 0.25\%\) \(\log{511.28} = \log{1.7} + \log{3} + \log{100} + 0.00432 * 0.25 = 2.70757 + 0.00108 = 2.70865\)\(511.28 = 500 + 2.256\% = 1000/2 + 2.256\%\) \(\log{511.28} = \log{1000} – \log{2} + 0.00432 * 2.256 = 2.69897 + 0.00974 = 2.70871\)
\(511.28 = 512 – 0.14\% = 2^9 – 0.14\%\) \(\log{511.28} = 9 * \log{2} – 0.00432 * 0.14 = 2.70927 – 0.00060 = 2.70867\)
The correct answer is \(2.7086588\) showing that we gained more accuracy by choosing a closer approximation; 510 and 512 gave more accurate answers than 500.
Summary:
With a small amount of practice, this method allows you to quickly generate logarithms accurate to 4 decimal places. For the reverse procedure (calculating \(10^x\) for any number \(x\)) see the article on calculating antilogarithms.
When combining the algorithms for logarithms and antilogarithms we have a general-purpose algorithm for all powers and roots.