This difficult method uses the properties of logarithms (with base 10) to present a general method for calculating any root or power. As examples, the following can be calculated effectively to an accuracy of about 4 digits using this method:
- \(2^{345}\)
- \(\pi^\pi\)
- \(\sqrt[7]{902.54}\)
For many such computations there are shortcuts and simpler methods. In particular there are better algorithms for square roots (accurate square roots method and fast square roots method) and for cube roots. For deeper roots a version of the Newton-Raphson method may be easier. For small integer powers such as \(41706^5\) direct multiplication is easier.
However for the most extreme roots and powers, this method works well. It is one successful method used in the surprise task of the Mental Calculation World Cup 2016, which was to compute:
\(\sqrt[4]{2016} – \sqrt[6]{2016} + \sqrt[7]{2016}\)
Mathematics:
For simplicity all logarithms and exponents are to base 10. 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 \(\log 1000\))
Mathematicians learn that the following are always true:
- \(\log (a*b) = \log a + \log b\)
- \(\log (\frac{a}{b}) = \log a – \log b\)
- \(\log (a^b) = b\log a\)
- \(\log (\sqrt[b]{a}) = \frac{\log a}{b}\)
- \(n = 10^{\log (n)}\)
So if we have a difficult root or power to calculate, we can first calculate its logarithm using rules (1)-(4) and then we can find the antilogarithm of that, which will be our final answer.
Preparation:
- You will need to know how to calculate logarithms and calculate antilogarithms
- You will need to have memorized enough values for logarithms (e.g. \(\log 2 = 0.30103\))
The Method:
The method is best illustrated by working through some examples, but the main stages always are:
- calculate the logarithm
- multiply (for powers) or divide (for roots) the logarithm
- calculate the antilogarithm of that to arrive at the final answer
Simple example for powers:
We seek to calculate \(2^{345}\)
Using rule (3), \(\log(2^{345}) = 345 * \log 2\)
We already memorized that \(\log 2 = 0.30103\) so this is \(345 * 0.30103 = 103.85535\)
Therefore using rule (5), \(2^{345} = 10^{103.85535}\)
We can simplify this with rule (1) to \(2^{345} = 10^{0.85535} * 10^{103}\)
Using the algorithm for estimating antilogarithms, we calculate that \(10^{0.85535} = 7.1672\)
Therefore our answer is \(7.1672 * 10^{103}\)
More difficult example for powers:
We attempt the impossible-looking \(\pi^\pi\). \(\log(\pi^\pi) = \pi * \log \pi\)
Using our algorithm for calculating logarithms we approximate \(\log \pi\) as follows:
\(\log{3.14159} = \log{\frac{22}{7} – 0.04\%} = \log{11} + \log{2} – \log{7} – 0.00432*0.04\)
\(\log{3.14159} = 1.04139 + 0.30103 – 0.84510 – 0.00017 = 0.49715\)Note: the 0.04% is from quickly approximating \(\frac{\frac{22}{7} – \pi}{\pi}\)
The most difficult part in this calculation is the blind multiplication of 3.14159 and 0.49715 to 5 decimal places:
\(\log(\pi^\pi) = 3.14159 * 0.49715 = 1.56184\)
Then again we use our method for antilogarithms to compute \(\pi^{\pi} = 10^{1.56184}\)
Try 36: \(\log{36} = 1.55630\) so \(\pi^{\pi} = 36 * 10^{0.00554}\)
\(0.00554/0.00432 = 1.28\) so \(\pi^{\pi} = 36 + 1.28\% = 36.46\)
Example for roots:
The same method works for any roots – except that we perform a division rather than a multiplication. As an example we will calculate \(\sqrt[7]{902.54}\)
Using the method for calculating logarithms, \(\log{902.54} = 2.95547\)
\(\log{\sqrt[7]{902.54}} = \frac{\log 902.54}{7} = \frac{2.95547}{7} = 0.42221\) \(\sqrt[7]{902.54} = 10^{0.42221}\)
Using the method for calculating antilogarithms, \(10^{0.42221} = 2.643\) and this is our answer.
Summary:
This is one of the most advanced algorithms used in mental calculation and can be used to solve almost any problem involving roots or powers. Note that unlike many mental calculation algorithms it only provides accuracy to around 4 digits.
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