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Niklas Oberhuber

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Estimating Logarithms

Published on 21st May 2025. 5 minutes long read.
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While reading through the fantastic book The Lost Art of Logarithms
by Charles Petzold I was nerd-sniped by a simple method of estimating
the logarithm of any number base 10. According to the book, it was
developed by John Napier (the father of the logarithm) about 1615.

    In french the natural logarithm is also called "le logarithm
    neperien" in reference to the mathematician.

The Method

We note that due to the nature of the logarithm (always referring to
base 10 from here one out), the logarithm of any number NNN is
approximately equal to the number of digits of NNN minus one. This is
quite easy to see when thinking about numbers between 100 and 1000
for example:

100<=N<1000 log[?](100)<=log[?](N)<log[?](100) 2<=log[?](N)<3\begin{align} 100 &
\leq N < 1000 \\\ \log(100) &\leq \log(N) < \log(100) \\\ 2 &\leq \
log(N) < 3 \end{align}100 log(100) 2 <=N<1000<=log(N)<log(100)<=log(N)<3
  

This approximation by itself might seem useless at first: knowing
that the logarithm of 5 is between 0 and 1 is pointless. But in
combination with the following property of logarithms:

log[?](ab)=b[?]log[?](a)\log(a ^ b) = b \cdot \log(a)log(ab)=b[?]log(a)

We can calculate the logarithm of any number with arbitrary precision
using the following this algorithm. We note #N\#N#N as the number of
digits of N minus one.

log[?](N)[?]#N log[?](N10)[?]#(N10) 10[?]log[?](N)[?]#(N10) log[?](N)[?]#(N10)10\begin
{align} \log(N) &\approx \# N \\\ \log(N ^{10}) &\approx \#({N^{10}})
\\\ 10 \cdot \log(N) &\approx \#({N^{10}}) \\\ \log(N) &\approx \frac
{\#({N^{10}})}{10} \end{align}log(N) log(N10) 10[?]log(N) log(N) [?]#N[?]#(
N10)[?]#(N10)[?]10#(N10)   

Increase the exponent from 100 to 1000 and you've added another digit
of precision.

Henry Briggs used this method to compute the logarithms of 2 and 7 to
the 14th digit. Calculating 210142^{10^{14}}21014 must have been
quite a task. He probably rounded the values a lot before
multiplying, as there is no use to the first few digits at those
scales.

Simply exchanging the complexity of calculating the logarithm with an
extremely tedious exponentiation isn't very useful, we are lacking
one more insight...

One more Trick

Fortunately, between the 17th century and today, mathematicians came
up with something that makes this task a lot easier: scientific
notation.

510[?]9.8[?]1065^{10} \approx 9.8 \cdot 10^6510[?]9.8[?]106 and 520=510[?]5105^
{20} = 5^{10} \cdot 5^{10}520=510[?]510. Thus 
520[?]9.8[?]106[?]9.8[?]106[?]9.8[?]9.8[?]1012[?]9.6[?]10135^{20} \approx 9.8 \cdot 10^
6 \cdot 9.8 \cdot 10^6 \approx 9.8 \cdot 9.8 \cdot 10^{12} \approx
9.6 \cdot 10^{13}520[?]9.8[?]106[?]9.8[?]106[?]9.8[?]9.8[?]1012[?]9.6[?]1013. This was
a lot easier to do than what it looked like before converting to
scientific notation. And instead of starting from 0 everytime, we can
use the results from our previous calculation and simpler additions
and multiplications to gain more precision.

Knowing that log[?](5)[?]#(9.6[?]1013)20[?]1320[?]0.65\log(5) \approx \frac{\#
(9.6 \cdot 10^{13})}{20} \approx \frac{13}{20} \approx 0.65log(5)[?]20#
(9.6[?]1013) [?]2013 [?]0.65, which isn't too far from the correct 0.690.69
0.69. To increase the exponent to 51005^{100}5100, which will gain us
even more precision, we will use the fact that 5100=
51010[?]9.810[?]10610[?]8.2[?]10695^{100} = 5^{10^{10}} \approx 9.8^{10} \
cdot 10^{6^{10}} \approx 8.2 \cdot 10^{69}5100=51010[?]9.810[?]10610[?]8.2[?]
1069.

By repeating these steps and keeping previous results, we only ever
have to multiply the mantissa (always smaller than 10) with itself
and the exponent by 2. After only 10 repetitions of exponentiating by
2 (a total of 20 rather trivial multiplications) we'll have 4 digits
of precision.

In Code

This is still a lot of work to do by hand and thus I went ahead and
wrote a little python script that starts from N10N^{10}N10 and
calculates the logarithm up to arbitrary precision.

import math
import decimal
decimal.getcontext().prec = 100

def get_scientific(num):
    num = decimal.Decimal(num)

    # Calculate length of the number in a 'naive' way, without using the log
    # Using log10 here would kind of defeat the point
    # Because the length never exceeds 10, counting them manually is always possible
    length = len(str(math.floor(num))) - 1
    mantissa = num / (10 ** length)

    return mantissa, length

def count_digits(num, precision: int):
    assert(precision > 0)

    mantissa, exponent = get_scientific(num ** 10)

    # Apply the trick here by doing the calculation iteratively
    for _ in range(precision - 1):
        # Use the properties of exponents
        # (m x 10^exp)^10 = m^10 x 10^(10 * exp)
        mantissa = mantissa ** 10
        exponent *= 10

        mantissa, new_exponent = get_scientific(mantissa)
        # Calculate the contribution of the new mantissa to the exponent and add it
        exponent += new_exponent

    return exponent

def logarithm(num, precision):
    # Count the number of digits
    n_digits = count_digits(num, precision)
    # Divide by the exponent
    result = n_digits / decimal.Decimal(10 ** precision)

    return f"{result:.{precision}f}"

For each additional digit of precision, it calculates Nprev10N^{\text
{prev}^{10}}Nprev10, where the scientific form of NprevN^{\text
{prev}}Nprev is already known, starting with N10N^{10}N10. This makes
the calculation a matter of exponentiating the mantissa by 10 and
multiplying the exponent by 10. [DEL:Rinse:DEL] keep the previous
results and repeat.

I'm not super happy with the implementation because of the decimal
import which could probably be made unnecessary by doing some string
manipulations.


I wrote this little post in order to remember this neat little trick
better and to maybe expose some people who wouldn't have read chapter
4 in the book to it. I can only recommend checking out the content
straight from the source, online and for free!

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