## 1.6 Short Examples

To get a feel for using logarithms in calculations, let’s perform a few examples. We will suppose that we have a table of logarithms already prepared, where the number $$x$$ is tabulated for the range $$1 \le x \le 10$$, sub-divided into intervals such as 10-ths, or 100-ths, say, and the logarithms $$\log x$$ are entered into the table with a certain accuracy, such as 5 decimal places, and are in the range $$\log 1 \le \log x \le \log 10$$, or equivalently, $$0 \le \log x \le 1.$$

For example, suppose we want to compute $$38900 \times 0.792$$. We start by re-writing the problem as follows:

$38900 \times 0.792 = 3.89 \times 10^{4} \times 7.92 \times 10^{-1} = 3.89 \times 7.92 \times 10^{3}.$

Then, taking the logarithm of this result,

$\log(3.89 \times 7.92 \times 10^3) = \log 3.89 + \log 7.92 + 3 \\$

Then, from a table of 5-place logarithms we find that

$\log(38900 \times 0.792) = \log 3.89 + \log 7.92 + 3 \\ = 0.58995 + 0.89873 +3 \\ = 1.48868 + 3 \\ = 0.48868 + 4.$

Next, we find the number between 1 and 10 that has the logarithm 0.48868. From our table of logarithms we should find that the number with logarithm 0.48868 is, to five significant figures: 3.0809. The FINAL result of our multiplication is hence $$3.0809 \times 10^{4}$$ = 30809. Check (via computer):

## [1] 38900 * 0.792  =  30808.8

As we can see, the answer is correct, though its accuracy from the logarithmic approach will depend upon the accuracy of the logarithm values used. Here, our answer with 5-place logarithms is good to 6 parts per million. Such accuracy might be important in some applications, but might be overkill in others – do you need to know the height of that tree in your backyard to within 50 microns?

As another example, let’s divide 245 by 5960. To start, write

$\frac{245}{5960} = \frac{2.45\times 10^2}{5.96 \times 10^3} = \frac{2.45}{5.96} \times 10^{-1}.$

Then, proceeding as before, we get

$\log(245/5960) = \log 2.45 - \log 5.96 -1 \\ = 0.38917 - 0.77525 -1 \\ = -0.38608 - 1 \\$

But our logarithm tables are for numbers that are greater than one, and hence with logarithms greater than zero. So, we take the above result and add 1 and subtract 1 to get

$\log(245/5960) = -0.38608 - 1 \\ = 1-0.38608 - 2 \\ = 0.61392 - 2$

which gives us a positive number between 0 and 1, plus an integer which in this case happens to be negative.

Then, from our logarithm table, we find that 0.61392 is the logarithm of 4.111. But the “-2” tells us to move the decimal point two places to the left. So, finally, we see that 245/5960 must be equal to 0.04111. By computer we find that 245/5960 = 0.041107.

Taking two numbers that we used earlier, let $$x$$ = 5827, and $$y$$ = 0.0365. The product is $$x\times y =$$ 212.6855, which can be verified by straightforward multiplication using pencil and paper, if you want to take the time. But now let’s compute this product using logarithms. Here, we’ll use a computer to do the work,4 where the function log10(a) below produces the Base 10 logarithm of the number a to a high degree of precision:

x = 5827
y = 0.0365
logx  = log10(x)
logy  = log10(y)

The output:

## [1] logx = 3.76544501809015 ; logy = -1.43770713554353 .
## [1] logx + logy = 2.32773788254662 .
## [1] 10^( 2.32773788254662 ) = 212.6855 .

The number with logarithm 2.32774 is 212.6855.

The computer example above illustrates well the usefulness of having values of logarithms accurate to many decimal places. Of course, with a computer one doesn’t need to perform multiplications using logarithms. But prior to computers, using tables of logarithms was a standard approach to perform multiplications and divisions with high accuracy, and was taught in schools until the late 1970s. In our next chapter we will see details of how to compute the logarithm of an arbitrary number.

Before electronic computing devices, one could have used a book of tables to look up the appropriate logarithms and then add them together to perform a “multiplication”. Then, one would take that result and find in the tables the number that had that value as its logarithm; this would be the final answer, with the computer (the one doing the computing) having to keep track of the powers of ten. This would be repeated as needed to multiply and/or divide a series of numbers.

The slide rule provides access to the values of logarithms (to a certain accuracy) and enables quick manipulations of them for mathematical calculations without the need to write down each of the steps and without looking up numbers in tables. The slide rule greatly sped up the process of performing long calculations, when the final results were only required to be accurate to a few digits. We will come back to this procedure when we get to Slide Rule ABC’s and D’s.

1. Throughout this text we are using the R programming language. See .↩︎