1950s calculators

Excerpt: This resource is designed to develop a sense of what logarithms are, how they can be used and the sort of values they take. A set of short video clips demonstrate the use of 'log tables' to perform a selection of multiplications and other calculations.

Log tables

Here is a question from a 1952 O-level exam paper:

Calculate, correct to three significant figures, $\frac{1{\cdot}152\times(3{\cdot}902)^3}{(5{\cdot}463)^2} .$

This is relatively straightforward to do using a calculator.

But electronic calculators only became commonly available from the late 1970s, so how did students of the 1950s answer a question like this?

They used logarithm tables (usually called “log tables”) where we would now use calculators, and in this resource we will explore how logarithms can help us to do calculations.

We’ll start with something simpler, and look at the exam question in a later section.

Use logarithm tables to evaluate, correct to three significant figures, $336\times6290 .$

We will do this in a series of steps.

Step 1

Can you write the numbers in this calculation ( $336$ and $6290$ ) in standard form?

Answer

336=3{\cdot}36\times10^2\quad\text{and}\quad6290=6{\cdot}29\times10^3

Step 2 (video)

In this video, you can see someone working out $\log 336$ using log tables.

Can you describe what they are doing and explain how it relates to standard form?

Explanation

The logarithm is found in two parts.

The table gives the digits of $\log{3{\cdot}36}$ (namely $5263$ ) and these go after the decimal point.

To get from $3{\cdot}36$ to $336$ we have to multiply by $10^2$ , so the whole number part of the logarithm is $2$ .

You can think of it as

\begin{align*} 336&=3{\cdot}36\times10^2\\ \log336&=\log{(3{\cdot}36\times10^2)}\\ &=\log{3{\cdot}36}+2\\ &=2+0{\cdot}5263 \end{align*}

Step 3 (video)

Can you work out $\log 6290$ using the log tables?

Check your answer in the following video:

Step 4 (video)

This clip shows how we can combine the two logs to work out the log of the answer.

(If you have already learnt about the laws of logarithms, you might like to try to do this step yourself before watching the video clip.)

Explanation

One of the laws of logarithms states that $\log ab = \log a + \log b$ so by adding the two logarithms we have found the logarithm of the answer.

For more on the laws of logarithms, you might like to look at the resource Proving the laws of logarithms.

Step 5 (video)

Finally, we convert from the logarithm to the final answer.

Can you explain the steps that they are doing? (The antilogarithm tables are on the second page of the

log tables

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Some examples for you to try

Work these out to three significant figures using the log tables and check your answers using a calculator.

$4620\times318$

$76{\cdot}4\times4{\cdot}12$

$197\div5{\cdot}63$

How have these logarithm tables made the calculations easier than they would be using other pencil-and-paper methods?

Add the difference

In the previous section we used log tables to find the logarithms of numbers given to three significant figures. You may have noticed that the tables have extra columns at the right. These are called “difference columns” and allow us to look up the logarithms of numbers given to four significant figures.

The following video shows someone calculating $28{\cdot}27{\times}185{\cdot}7$ .

Practise using the log tables to calculate $173{\cdot}5{\times}6{\cdot}384$ and check your answer using an electronic calculator.

Can you apply the other laws of logarithms to calculate the following using log tables?

$\dfrac{4866}{343{\cdot}9}$

$1{\cdot}046^8$

$\sqrt[3]{399500}$

You can check your working and your solutions here.

Solution 1

$\dfrac{4866}{343{\cdot}9}$

Solution 2

$1{\cdot}046^8$

Solution 3

$\sqrt[3]{399500}$

Why would it be inappropriate to give the final answers to 4 significant figures?

1952 exam question

Now here is the full version of the 1952 exam question we started with earlier.

Use logarithms to calculate, correct to three significant figures, $\frac{1{\cdot}152\times(3{\cdot}902)^3}{(5{\cdot}463)^2} .$

The area of a square field is $5{\cdot}355$ hectares. Find, correct to the nearest metre, the length of a diagonal of the field. [1 hectare = 100 ares; 1 are = 1 square dekametre.] [Editorial note: 1 dekametre = 10 metres.]

UCLES O level Elementary Mathematics Alternative A: Arithmetic, Question 7, 14 July 1952 (slightly amended)

Solution ( i )

Solution ( ii )

One final twist

In this resource we have dealt only with the logarithms of numbers greater than $1$ . Can you see the difficulty with using log tables to do calculations with numbers such as $0{\cdot}0283$ ?

Bar notation

First, we write the number in standard form: $0{\cdot}0283=2{\cdot}83\times10^{-2}$ so $\log 0{\cdot}0283=-2+\log 2{\cdot}83=-2+0{\cdot}4518 .$

If we were to work this out as $-2+0{\cdot}4518=-1{\cdot}5482$ , we would lose the clarity that the power of $10$ is $-2$ . So instead we write the number as $\bar2{\cdot}4518$ (called bar notation). Arithmetic with such quantities is carried out separately on the whole number part (called the characteristic) and the decimal part (called the mantissa) as illustrated here. Use of bar notation usually makes calculations easier.