 # Logarithms

 From: Nick Jennings Date: 15 Feb 1999 Subject: What are logarithms used for? I'm doing A-level Pure Maths and we have spent several lessons on logarithms. None of us really understand what we are doing, it just seems like a lot of work for the sake of it. Do logarithms have any applications in the real world, or is it just to keep maths teachers in a job???

### Maths Help suggests:

When logarithms were invented almost 400 years ago (by a Scotsman,
John Napier), it was claimed that they "doubled the life of astronomers".
At that time, astronomers were among the most influential scientists,
and they found that using logarithms they could do calculations twice as
quickly as they previously could.

Nowadays, we use pocket calculators rather than logarithms to carry out
long multiplications and divisions and other calculations. But there are
many other areas where the use of logarithms is essential to help solve
mathematical problems. You will meet quite a few if you stick with your
A-level course!

For now, we will present one important applications. Logarithms are
necessary to solve problems with unknown powers.

EXAMPLE: Find the value of x for which 2x = 95

We could attack this problem by trial and error, noting that:
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
27 = 128
So the value of x which makes 2x = 95 is between 6 and 7,
and experimenting with your calculator will enable you to get closer
and closer to the answer by trial and error. [Try it.]

However, trial and error is not very efficient. Here is a better way:
Start with the equation 2x = 95
Take logs of both sides: log2x = log95
Apply "Rule 3" of logs: x.log2 = log95
Divide through by log2: x = log95/log2
Evaluate on calculator: x = 6.569855608......

You ask for a practical application of this in the "real world".
Well, suppose you invest a sum of money at a fixed rate of 7% interest
per annum. Then at the end of each year the money is worth 1.07 times
what it was at the begining of the year (100% plus the extra 7% makes
107% = 107/100 = 1.07 of the original). So after n years it is worth
(1.07)n times what it was originally. How long will it take
until your investment doubles in value?
To answer this, you need to find the value of n for which (1.07)n = 2
[Check yourself, using the method shown above, that n = 10.24....
meaning that the capital should remain invested for 10 years 3 months.
]

Calculations like this are carried out all the time in the financial world.
It is not just maths teachers who are kept in a job!

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