Sunday, 15 December 2019

Fun With Numbers

This week I have been mostly thinking about natural logarithms.

I'm not sure if there's any practical purpose to this in everyday life, but it means expressing a number as a certain power of e.

e (or e^1) = 2.7 1828 1828 459 (I have put the spaces in because this is how you memorise it to 12 decimal places). The natural log of e = 1.

e^2 = 7.389, so the natural log of 7.389 = 2.0, and so on.

To approximate the natural log of any number up to 1,000, you just need to memorise two tables, I've rounded all the numbers a bit because this is only an approximation:

Table 1
Natural log of 2.7 = 1
Natural log of 7.4 = 2
Natural log of 20 = 3
Natural log of 55 = 4
Natural log of 148 = 5
Natural log of 403 = 6

Table 2
25% = 0.22
50% = 0.41
75% = 0.56
100% = 0.69
125% = 0.81
150% = 0.92

You can memorise more increments if you wish, but it's just as easy interpolating between them.

So... how do you work out the natural log of, for example, 80, to one decimal place?

The highest number from Table 1 that is lower than 80 is 55, so the first part of your answer is 4.

Then subtract 55 from 80 = 25. That's 'a bit less than' 50% of 55. You look in Table 2, and the next part of your answer is 'a bit less than' 0.41, call it 0.4.

Add the two parts together, the natural log of 80 is 4.4.

Check on calculator, yup, the answer is 4.382. You can put in a bit more effort and interpolate between 0.22 and 0.41 to get close to the 0.38 bit.

Let's do 100.

The first part of the answer is also 4, from Table 1.

100 - 55 = 45, which is 'a bit more than' 75% of 55, so look up 75% in Table 2, the next part of your answer is 'a bit more than' 0.56, so round it up to 0.6

4 + 0.6, the natural log of 100 is about 4.6.

Check on calculator, the answer is 4.605.
After half an hour's practice, you can get most to the nearest decimal place, two decimal places if you put in the effort and strike lucky. It's a bit more challenging than playing Candy Crush or something.


View from the Solent said...

I agree that natural logs have no everyday purpose, but large sections of maths and fizzics would collapse without e and its logs.

And Euler's identity: e^(i x pi) = -1
is just beautiful

Mark Wadsworth said...

VFTS, that formula is maths porn, same as e=mc^2 is physics porn.

Bayard said...

VFTS, I prefer the form e^(iπ)+1=0, as it then contains all the major numbers

James Higham said...

This week I have been mostly thinking about natural logarithms.

As one does of course.

Mark Wadsworth said...

JH, have you never wondered what the heck they are all about? Well, now we know.