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Example 4. Required the logarithm of 5.
10=1.000000000 subtract the logarithm of 2=0.301029995
the remainder is the logarithm of 550.698970005 Example 5. Required the logarithm of 6. 6=3X2, therefore to the logarithm of 3=0.477121254
add the logarithm of 2=0.30 1029995
their sum= - logarithm of 6=0.778151249 Example 6. Required the logarithm of 8. 8=23, therefore multiply the logarithm of 2=0.301029995
The product = logarithm of 8=0.903089985 Example 7. Required the logarithm of 9. 9=3", therefore the logarithm of 3=0.477121254 being multiplied by
the product = logarithm of 9=0.954242508 Example 8. Required the logarithm of 7.
Here the next less number is 6, and 7+6=13=A, and A?=169.
of 3 and 4 of 14
of 7 and 2. The logarithm
of 15 is equal to the sum of the of 3 and 5. of 16 logarithms
of 4 and 4. of 18
of 3 and 6. of 20
of 4 and 5. The logarithms of the prime numbers 11, 13, 17, 19, &c. being computed by the foregoing general rule, the logarithms of the intermediate numbers are easily found by composition and division. It may however be observed, that the operation is shorter in the larger prime numbers; for when any given number exceeds 400, the first quotient being added to the logarithm of its next lesser number, will give the logarithm sought, true to eight or nine places; and therefore it will be very easy to examine any suspected logarithm in the Tables.
For the arrangement of logarithms in a table, the method of finding the logarithm of any natural number, and of finding the natural number corresponding to any given logarithm therein,likewise for particular rules concerning the indices, the reader will consult Table 1, with its explanation at the end of this treatise.
MULTIPLICATION. Two or more numbers being given, to find their product by
Having found the logarithms of the given numbers in the table, add them together, and their sum is the logarithm of the product; which logarithm being found in the table, will give a natural number, that is, the product required.
Whatever is carried from the decimal part of the logarithm is to be added to the affirmative indices, but subtracted from the negative. Likewise the indices must be added together when they are all of the same kind, that is, when they are all affirmative, or all negative; but when they are of different kinds, the difference must be found, which will be of the same denomination with the greater.
Example 1. Required the product of 86.25 multiplied by 6.48.
Log. of 86.25 : 1.935759
Product 558.9 2.747334* * For the method of finding the natural number, answering to the sum of the logarithms, the reader will consult Table 1, at the end of this treatise.
Example 2. Required the product of 46.75 and .3275.
Log. of 46.75 1.669782
Product 15.31+=1.184993 Here the + 1 that is to be carried from the decimals cancels the -1, and consequently there remains 1 in the upper line to be set down.
Example 3. Required the product of 3.768, 2.053, and .007693.
3.768 0.576111 Log. of
2.053 0.312389 Log. of .007693 =43.886096
Product = .05951+=-2.774596 In this example there is 1 to carry from the decimal part of the logarithms, which, subtracted from —3, the negative index, leaves 2, the index of the sum of the logarithms, and is negative.
Example 4. Required the product of 27.63, 1.859, .7258, and 0.3591.
Log. of 27.63 1.441381
Product nearly 1.339 0.126692 In this example there is 2 to carry from the decimal part of the logarithms, which added to 1, the affirmative index, makes 3, from this take 3, the sum of the negative indices, the remainder is 0, which is the index of the sum of the logarithms. 5. Required the product of 23.14 and 50.62, by logarithms.
Ans. 117.1347 6. Required the product of 3.12567, .02868, and .12379, by logarithms.
Ans. .01109705 7. Required the product of .1508, .0139, and 756.9, by logarithms.
Ans. 1.586553 8. Required the product of 637.8 and 89.27, by logarithms.
Ans. 56936.406 9. Required the product of 14 and 8.45, by logarithms.
DIVISION. Two numbers being given, to find how many times one is con
tained in the other by Logarithms.
From the logarithm of the dividend subtract the logarithm of the divisor, and the remainder will be the logarithm whose corresponding natural number will be the quotient required.
In this operation, the index of the divisor must be changed from affirmative to negative, or from negative to affirmative; and then the difference of the affirmative and negative indices must be taken for the index to the logarithm of the quotient. Likewise when 1 has been borrowed in the left-hand place of the decimal part of the logarithm, add it to the index of the divisor, if affirmative; but subtract it, if negative; and let the index thence arising be changed and worked with as before. Example 1. Divide 558.9 by 6.48.
Log. of 558.9 : 2.747334
Quotient 85.25 1.935759 Here, the 1 to be taken from the decimals is taken as which when added to 2, the index of the dividend, leaves 1 for the index of the quotient; that is, 2—1=1. Example 2. Divide 15.31 by 46.75.
Log. of 15.31 1.184975
Quotient=.3275 -1.515193 Here, the 1 to be taken from the decimals is added to 1, the index of the divisor makes 2; this with its sign changed is —2, from which subtracting 1, the index of the dividend, the remainder is -1, which is negative, because the negative index is greater. Example 3. Divide .05951 by .007693.
Log. of .05951 2.774599
Quotient 7.735 = 0.888494 Here, the 1 to be taken from the decimals is subtracted from-3, which leave —2, this changed is +2;
and this added to
-2, the other index, gives 242=0.
Example 4. Divide .6651 by 22.5.
Log. of .6651=-1.822887
Quotient = .02956 =-2.470704 Here, +1 in the lower index, is changed into -1, and this added to -1, the other index, gives -1-1, or -2, the index of the result.
5. Required the quotient of 125 divided by 1728, by logarithms.
Ans. .0723379 6. Divide 1728.95 by 1.10678, by logarithms.
Ans. 1562.144 7. Divide .067859 by 1234.59, by logarithms.
Ans. .0000549648 8. Divide .7438 by 12.9470, by logarithms. Ans. .057449 9. Divide .06314 by .007241, by logarithms. Ans. 8.71979
Or the Rule of Proportion in Logarithms.
Having stated the three given terms according to the rule in common Arithmetic, write them orderly under one another, with the signs of proportion; then add the logarithms of the second and third terms together, and from their sum subtract the logarithm of the first term, and the remainder will be the logarithm of the fourth term, or answer.
Or,-add together the arithmetical complement of the logarithm of the first term, and the logarithms of the second and third terms ; the sum, rejecting 10 from the index, will be the logarithm of the fourth term, or term required.
N.B. The arithmetical complement of a logarithm is what it wants of 10,000000, or 20,000000, and the easiest way to find it is to begin at the left-hand, and subtract every figure from 9, except the last, which should be taken from 10; but if the index exceed 9, it must be taken from 19.-It is frequently used in the rule of Proportion and Trigonometrical calculations, to change subtractions into additions.
* When the index is negative add it to 9, and subtract as before. And for every arithm ical complement that is added, subtract 10 from the last sum of the indices.