Quick calculations for extremely large numbers

A guest post by Nandeesh H.N. of Kolkata

Quick calculations with a few logarithms

If you can remember a few logarithms, you can do many calculations quite easily without the aid of calculators or computers.

Try to remember the logarithms of just seven numbers:

Log 2 = 0.30, log 3 = 0.48, log 7 = 0.85, log 11= 1.04, log 13= 1.11, log 17 = 1.23 and log 19=1.28.

The logarithm of a composite number is equal to the sum of the logarithms of its prime factors; you can formulate the following table of logarithms:

NumberLogarithmNumberLogarithm
20.30111.04
30.48121.08
40.60131.11
50.70141.15
60.78151.18
70.85161.20
80.90171.23
90.95181.26
191.28

Note: Logarithm of 11 is 1.04. It means log 1.1 = 0.04. Log 5 = log (10/2) = log10 – log 2.

With the above figures in hand you can do a lot of calculations quite easily.

1. What is 2 raised to power 37?

Logarithm of 2 is 0.30.

37 times 0.30 is 11.10

Since logarithm of 1.2 is 0.08 and logarithm of 1.3 is 0.11, antilog of 0.10 is about 1.26.

So antilog of 11.10 is 1.26 * 10^11.

That is 2^37 = 1.26 * 10^11.

2. What is the 31st root of a 35 digit number?

This can be answered even without knowing the number.

Log of 35th digit number is 34. …

Log of 31st root   = 34. …./ 31

This lies between 34/31 and 34.99/31   or between 1.09 and 1.13.

From the above table, we can see that in this interval, logarithm of 13 is 1.11.

So, 31st root of a 35 digit number is 13.

3. What is the 64th root of a 20 digit number? Answer is 2.

4. What is the cube root of say 74567?

Log of 74567 = 4.873  (interpolating between log 7 =0.85 and log 8 = 0.90).

4.873 / 3 = 1.624.

Antilog of 0.624 is 4.24 (because log 4 = 0.60 and log 5 = 0.70).

So, cube root of 74567 is about 42.4

This post is contributed by Nandeesh Nagarajaia. He is a Chemical Engineer who did his B.Tech from NIT Suratkal.  He is now in IT field as Assistant General Manager(Systems) in Hindustan Copper Limited. He love Maths and enjoy teaching Maths to his sons.

On behalf of all the QuickerMaths.com users, I  am highly grateful for his contribution.

4 Comments
  1. Charchika
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