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Hence the log. of 8.9768713 is .953125.

And, by proceeding in this manner, it will be found, after 25 extractions, that the logarithm of 8.9999998 is .9542425, which may be taken for the logarithm of 9, as it differs from it so little that it may be considered as sufficiently exact for all practical purposes.

And in this manner were the logarithms of all the prime numbers at first computed.

RULE 2. When the logarithm of any number (n) is known, the logarithm of the next greater number may be readily found from the following series, by calculating a sufficient number of its terms, and then adding the given logarithm to their sum. Log. (n + 1) = log. n + M'

§

+ 2n+1 + 3(2n +1)

3(2n +1) 5(2n + 1). 1 1.

1 +

+

&c. 7(2n + 1)" 9(2n +1) 11(2n + 1)!1'

1

1

1

+

Or,

A

Зв

50

7D

9E

M' Log. (n + 1) = log. n +

+ 22n+1 + 3(2n+1)2 + 5(2n +1) + +

&c. 7(2n+1) 9(2n+1) 11 (2n + 1)?

Where A, B, C, &c., represent the terms immediately preceding those in which they are first used, and m' = twice the modulus = .8685889638

*

EXAMPLES.

1. Let it be required to find the common logarithm of the number 2.

Here, because n + 1 2, and consequently n=1 and 2n + 1 =3, we shall have 8685889638

.289529654 (A) 2n + 1

3 .289529654

.010723321 (B) 3(2n +1)% 3.32

M'

А

* It may here be remarked, that the difference between the logarithms of any two consecutive numbers is so much the less as the numbers are greater; and consequently, the series which comprises the latter part of the above expression will in that case converge so much the faster.

1 Thus, log. n and log. (n-+-1), or its equal, log. n+log. (1+), will obviously differ but little from each other when n is a larger number.

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A

Which logarithm is true to the last figure inclusively.

2. Let it be required to compute the logarithm of the number 3.

Here, since n+1=3, and consequently, n=2, and 2n+1=5, we shall have M' .868588964

1.173717793 (A) 2n + 1

5
.173717793

.002316237 (B)
3 (2n +1) 3.52
3B
3 x .002316237

=.000055590 (c)
5 (2n+1)

5.59
50
5 X .000055590

.000001588 (D)
7 (2n + 1)?

7.52
70
7 x .000001588

.000000050 (E)
9 (2n + 1)?

9.52
9E
9 x .000000050

.000000002 (F)
11 (2n +- 1)

11.52

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Which logarithm is also correct to the nearest unit in the last figure.

And in same way we may proceed to find the logarithm of any prime number.

Also, because the sum of the logarithms of any two numbers gives the logarithm of their product, and the difference of the logarithms the logarithm of their quotient, &c.; we may readily compute, from the above two logarithms, and the logarithm of 10, which is 1, a great number of other logarithms, as in the following examples: 3. Because 2 X 2= 4, therefore

.301029995

log. 2

Mult. by 2

2

gives log. 4 .602059990

4. Because 2 X 3 == 6, therefore 2

.301029995
to log. 2)
add log. 3 .477121255

gives log. 6 .778151250

5. Because 23 = 8, therefore log. 2 .301029995

Mult. by 3

3

gives log. 8 .903089985

6. Because 32 = 9, therefore log. 3 .477121255

Mult. by 2

2

gives log. 9 .954242510

1.000000000

7. Because = 5, therefore

from log. 10

take log. 2

.301029995

gives log. 5 .698970005

8. Because 3 x 4 = 12, there

.477121255 fore to log. 3

add log. 4 .602059991

gives log. 12 1.079181246

And thus, by computing, according to the general formula, the logarithms of the next succeeding prime numbers 7, 11, 13, 17, 19, 23, &c., we can find, by means of the simple rules before laid down for multiplication, division, and the raising of powers, as many other logarithms as we please, or may speedily examine any logarithm in the table.

MULTPILICATION BY LOGARITHMS.

TAKE out the logarithms of the factors from the table, and add them together; then the natural number, answering to the sum, will be the product required.

Observing, in the addition, that what is to be carried from the decimal part of the logarithms is also affirmative, and must, therefore, be added to the indices, or integral parts, after the manner of positive and negative quantities in algebra.

Which method will be found much more convenient, to those who possess a slight knowledge of this science, than that of using the arithmetical complements.

EXAMPLES.

1. Multiply 37.153 by 4.086, by logarithms. Nos.

Logs. 37.153

1.5699939 4.086

0.6112984

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2. Multiply 112.246 by 13.958, by logarithms. Nos.

Logs. 112.246.

2.0491709 13.958

1,1448232

[blocks in formation]

3. Multiply 46.7512 by .3275, by logarithms. Nos.

Logs. 46.7512.

1.6697928 .3275

ī.5152113

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Here, the + 1, that is to be carried from the decimals, cancel the

- 1, and consequently there remains 1 in the upper line to be set down.

4. Multiply, .37816 by .04782, by logarithms.
Nos.

Logs.
.37816

1.5776756 :04782

2.6796096

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Here the + 1, that is to be carried from the decimals, destroys the -1, in the upper line, as before, and there remains the 2 to be set down. 5. Multiply 3.768, 2.053, and .007693, together. Nos.

Logs. 19.768

0.5761109 2.053

0.3123889 .007693

3.8860997

[blocks in formation]

Here the + 1, that is to be carried from the decimals, when added to

- 3, makes 2 to be set down.
6 Multiply 3.586, 2.1046, 8372, and 0294, together.
Nos.

Logs.
3.586

0.554610 2.1046

0.323170 .8372

1.922829

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Here the + 2, that is to be carried, cancels the -2, and there remains the 1 to be set down. 7. Multiply 23.14 by 5.062 by logarithms.

Ans. 117.1347. 8. Multiply 4.0763 by 9.8432, by logarithms.

Ans. 40.12383. 9. Multiply 498.256 by 41.2467, by logarithms.

Ans. 20551.41. 10. Multiply 4.026747 by .012345, by logarithms.

Ans. 0497102. 11. Multiply 3.12567, .02868, and .12379, together, by logarithms.

Ans. .01109705. 12. Multiply 2876.9, .10674 .098762, and .0031598, by logarithms.

Ans. .0958299.

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