5. Interpolation of Logarithms.

In all that precedes we have used only logarithms of numbers containing not more than 4 significant digits. But in practice numbers of more than four figures have to be used. To find the logarithms of such numbers the process of interpolation is necessary. This process is one of simple proportion, which can be seen from the following example.

To find log. 1167.23.

The table gives the logarithms of 1167 and of 1168, which we find to be as follows:

Now the number of which we wish to find the logarithm being between these numbers, its logarithm is between these logarithms; that is, it is equal to 3.067 07 plus a fraction less than .000 37.

Since the difference 37 corresponds to the difference of unity in the two numbers, we assume that the quantity to be added to the logarithm bears the same proportion to .23 that 37 does to unity. We therefore state the proportion

1.23 37 increase required.

The solution of this proportion gives .23 x 37 8.51, which is the quantity to be added to log 1167 to produce the logarithm required. The result is 3.067 155 1.

But our logarithms extend only to five places of decimals, while the result we have written has seven. We therefore take only five places of decimals. If we write the mantissa 3.067 15, the result will be too small by .51. If we write 3.067 16, it will be too great by .49. Since the last result is nearer than the first, we give it the preference, and write for the required logarithm

log 1167.23 = 3.067 16.

We thus have the following rule for interpolating:

Take from the table the logarithm corresponding to the first four significant digits of the number.

Considering the following digits as a decimal fraction, multiply the difference between the logarithm and the next one following by such decimal fraction.

* In this multiplication we have used a decimal point to mark off the fifth order of decimals. This is a convenient process in all such computations.

This product being added to the logarithm of the table will give the logarithm required.

The whole operation by which we have found log 1167.23 would then be as follows:

The products for interpolation, 7.4 and 1.11, may be found by multiplying by the fifth and sixth figures of the number separately.

To facilitate this multiplication, tables of proportional parts are given in the margin. Each difference between two logarithms will be readily found in heavy type not far from that part of the table which is entered, and under it is given its product by .1, .2, etc., . . .9. We therefore enter this little table with the fifth figure, and take out the corresponding number to be added to the logarithm. Then if there is a sixth figure, we enter with that also and move the decimal one place to the left. We then add the two sums to the logarithm.

6. Labor-saving Devices.

In using a table of logarithms, the student should accustom himself to certain devices by which the work may be greatly facilitated.

In the first place it is not necessary to take the whole difference between two consecutive logarithms. He has only to subtract the last figure of the preceding logarithm from the last one of the following, increased by 10 if necessary, and thus find the last figure of the difference.

The nearest difference in the margin of the table having this same last figure will always be the difference required.

Example. If the first four figures of the number are 1494, instead of subtracting 435 from 464 we say 5 from 14 leaves 9, and look for the nearest difference which has 9 for its last figure. This we readily find to be 29, at the top of the next page.

NOTE. In nearly all cases the difference will be found on the same page with the logarithm. The only exception is at the bottom of the first page, where, owing to the number of differences, they cannot all be printed.

In the preceding examples we have written down the numbers in full, which it is well that the beginner should do for himself. But after a little practice it will be unnecessary to write down anything

but the logarithm finally taken out. The student should accustom himself to take the proportional parts mentally, adding them to the logarithm of the table and writing down the sum at sight. The habit of doing this easily and correctly can be readily acquired by practice.

Exercises. Find the logarithms of

As a precaution in taking out logarithms, the computer should always, after he has got his result, look into the table and see that it does really fall between two consecutive logarithms in the table.

If the fraction to be interpolated is nearly unity, especially if it is equal to or greater than 9, it will generally be more convenient to multiply the difference of the logarithms by the complement* of the fraction and subtract the product from the logarithm next succeeding. The following are examples of the two methods, which may always be applied whether the fraction be large or small:

Example 1. log 1004.28 = log (1005 — .72).

* By the complement or arithmetical complement of a decimal fraction is here meant the remainder found by subtracting it from unity or from a unit of the next order higher than itself.

Thus:

co. .723.327

co. .1796.8204

co. .9932.0068.

7. To find the Number corresponding to a given

Logarithm.

The reverse process of finding the number corresponding to a given logarithm will be seen by the following example:

To find the number of which the logarithm is 2.027 90.

Entering the table, we find that this logarithm does not exactly occur in the table. We therefore take the next smaller logarithm, which we find to be as follows:

log 1066 = 2.027 76.

Subtracting this from the given logarithm we find the latter to be greater by 14, while the difference between the two logarithms of the table is 40. We therefore state the proportion

40 14: 1 to the required fraction.

The result is obtained by dividing 14 by 40, giving a quotient .35. The required number is therefore 106.635. It will be remarked that we take no account of the characteristic and position of the decimal until we write down the final result, when we place the decimal in the proper position.

The table of proportional parts is used to find the fifth and sixth figures of the number by the following rule:

If the given logarithm is not found in the table, note the excess of the given logarithm above the next smaller one in the table, which call 4.

Take the difference of the two tabular logarithms, and find it among the large figures which head the proportional parts.

That proportional part next smaller than 4 will be the fifth figure of the required number.

Take the excess of 4 above this proportional part; imagine its decimal point removed one place to the right, and find the nearest number of the table.

This number will be the sixth figure of the required number. Example. To find the number of which the logarithm is 2.193 59. Entering the table, we find the next smaller logarithm to be .193 40. Therefore 4 = 19.

Also its tabular difference = 28.

Entering the table of proportional parts under 28, we find 16.8 opposite 6 to be the number next smaller than 19 the value of 4. Therefore the fifth figure of the number is 6.

The excess of 19 above 16.8 is 2.2. Looking in the same table for the number 22, we find the nearest to be opposite 8.

Therefore the fifth and sixth figures of the required number are 68. Now looking at the log .193 40 and taking the corresponding number, we find the whole required number to be

156 168.

The characteristic being 2, the number should have three figures before the decimal point. Therefore we insert the decimal point at the proper place, giving as the final result 156.168.

8. Number of Decimals necessary.

In the preceding examples we have shown how with these tables the numbers may be taken out to six figures. In reality, however, it will seldom be worth while to write down more than five figures. That is, we may be satisfied by adding only one figure to the four found from the table. In this case, when we enter the table of proportional parts, we take only the number corresponding to the nearest proportional part.

To return to the last preceding example, where we find the number corresponding to 2.193 59. We find under the difference 28 that the number nearest 19 is 19.6, which is opposite 7.

Therefore the number to be written down would be 156.17.

In the following exercises it would be well for the student to write six figures when the number is found on one of the first two pages of the table and only five when on one of the following pages. The reason of this will be shown subsequently.

Here we have a case in which the half of an odd number is required. We might have written the last logarithm 0.088 045, but we should then have had six decimals, whereas, as our tables only give five decimals, we drop the sixth. If we write 4 for the fifth figure it will be too small by half a unit, and if we write 5 it will be too large by half a unit. It is therefore indifferent which figure we write, so far as mere accuracy is concerned.