In this article we learn to calculate \(10^x\) for any value of \(x\), often called the antilogarithm of \(x\).
This is required for the method to calculate difficult roots and powers.
To understand this page you will need to have understood the method for calculating logarithms, because the antilogarithm method is the reverse of that.
Method:
- Using the memorized values for various logarithms, find a number whose logarithm is close to the target number
- Find the error between the target and this logarithm and if necessary return to the previous step to find a better candidate logarithm
- When you have selected a sufficiently close logarithm and have calculated its error, compute the correction factor, which is approximately 1% for every 0.00432 in the error
Example #1:
We wish to calculate \(10^{1.23456}\)
We already memorized \(\log{1.7} = 0.23045\)
Therefore \(10^{1.23456} = 10 * 1.7 * 10^{0.00411}\) because \(10^1 = 10\) and \(1.23456 – 1 – 0.23045 = 0.00411\)
\(\frac{0.00411}{0.00432} = 0.95\) so \(10^{1.23456} = 17 + 0.95\% = 17.1615\)
The correct answer is \(17.16169\) so our answer was correct to 5 digits. Performing a more accurate division \(\frac{0.00411}{0.00432} = 0.951\) would have given a more accurate answer of \(17.16167\).
Example #2:
We attempt to calculate \(10^{2.71828}\) and commence by searching for logarithms close to \(0.71828\):
- \(\log 5 = 1 – \log 2 = 0.69897\) (so 5 is too small)
- \(\log 512 = 9 * \log 2 = 2.70927\) (so 5.12 is too small)
However 5.12 brings us about halfway from 5 to the target logarithm \(0.71828\) so we should try somewhere around 5.24… we could try 5.2 or 5.25:
- \(\log 52 = \log 13 + 2 * \log 2 = 1.11394 + 0.60206 = 1.71600\)
This is close enough, so we know that \(10^{2.71828} = 520 + 10^{0.00228}\)
\(\frac{0.00228}{0.00432} = 0.53\) so \(10^{2.71828} = 520 + 0.53\% = 522.76\)
The correct answer is \(522.73\) so our answer is correct to 4 digits.
Summary:
This method efficiently computes antilogarithms although requires some intuition to perform efficiently. For the reverse procedure, see the article on calculating logarithms.
When combining the algorithms for logarithms and antilogarithms we have a general-purpose algorithm for all powers and roots.
If you have any questions or would like specialist coaching from myself, you’re welcome to contact me here.