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1 + 2 Theorem 6.- The logarithm of the number is equal to the
2x3 2x5 2x? 2.39 sum of the series M (2x + + +
&c.) 3 5 7 9
2x2 For, if in the serical equation, log. (1 + x)= M 2x - +
2 2.33 2x4 2x5
+ &c. we substitute - x for x, then log. (1-2)= 3
74 xs M (-X
&c.); subtract the latter equa2 3
5 tion from the former, and we shall have the equation log.(1 + x)
2x7 - log. (1-x) = M (2x +
+ &c.); but, by 5 17
1 + x Theorem 2, log. (1 + x) - log.(1-x) = log. ; therefore 1 + x
2xs =M(2x +
+ + &c.)
3 5 7 Theorem 7.-The logarithm of the number 2 + v is equal to the
2v3 sum of the series, log. 2 + M
+ (22 + x)?
3 (2z + v)' &c.) 5 (22 + v) 1 x
+ + 1
3 5 7 &c.) we substitute for x, we shall have the equation log.
2x + v
+ &c. 3(2x + v)
5(2z + us % + but because the log. = log. (x + v) – log. E, therefore log.
Qus '% + v) log. ? = M
+ 11(22 + 0' 3(2 + v) 5(22 + os to &c.
; or, by transposing log. 2, we shall have log. (: + v) =
205 log. 2 + M
&c. 1(22 + 1)' 3/2: + 3 5(22 + v)s Q.E.D. Cor.— If v= 1 and n=z + 1, then the log. (z + 1) = log.n and log.
SCHOLIUM.—The series announced in Theorem 5 is of very little use in the calculation of sogarithms, owing to the very slow degree of the convergency of its terms. But, in large numbers, the series announced in the last Theorem, or its Corollary, must be used; for, having the logarithm of a low number, that of a higher one may be found by the scries +
PROBLEM.-To find the logarithm of a low number by the 1 + x
25 x? equation log = M(x +
+ + &c.) according 3
7 to the Briggian system.
1 + x Put equal to the proposed number, which let it be n, then
1 1 + x
n-1 = n, therefore a =
For substitute 1n +
n + 1 in the series, and, instead of M, substitute 2 X .4342944 3 = .86858196, and we shall have the equation
(n-1)3 log. n = .86858896
3(n+1)3 5(n + i) Reduce the vulgar fractions to decimals, add the decimals of the terins together, and the sum will be the logarithm of the number required; or, according to the following
Rule.-Multiply the uumber •86858896 by the numerator of the value of x, divide the product by the denominator of the same value.
Multiply the quotient by the square of the numerator, and divide the product by the square of the denominator.
Continue this last operation of multiplying and dividing the last quotient by the squares of the terms of the fraction, keeping the last or right hand figure under the last of that above, until the last quotient be very small.
Divide the first quotient by 1, the second by 3, the third by 5, &c. until the last divisor canuot be contained in the last of the preceding quotients, keeping the right-hand figure of any quotient under that above, as before; then the sum of the last quotient will be the lo. garithm of the number required.
Example 1.–Find the logarithm of the number 2. Here n = 2, therefore in
-1=2-1=1, and n+1= 2 + 1= 3; whence (n − 1) =1, and (n + 1) = 9. The operation will then be as follows:
490 9,490 ; 11
4: 54 • 13 =
sum .3010299 Whence the log. 2 = .30102996 nearly.
PROBLEM.---To find the logarithm of a number, the logarithm of the next less number being given.-- In the equation log. n = 2
2 n M
3(2n-1 5(2n-1)S Then proceed according to the following
Rule.- For n substitute the number, the logarithm of which is required.
Divide the number .868589, first, by the value of 2n - 1, then. the quotient by (2n — 1)?, and so on.
Proceed in the same manner, always dividing the last quotient by (2x - 1).
Divide the quotients, as they succeed each other, by the corre sponding numbers of the series of odd numbers, 1, 3, 5, &c.
To the sum of the quotients add the logarithm of the next lower number, and the total sum wili be the logarithm of the number required.
Example. Find the logarithm of 19, the logarithm of 18 being 1.255273. Here n = 19; whence 2n · 1 = 38 – 1= 37, and 372 = 1369, whence the operation
.868589 • 37 = .023475
.023475 • 1369 = 17 Again, .023 +75 +
PROBLEM 1.-To find the logorithm oj any number by the table.
1. When the given number is less than 100.—Look for the given num. ber under N in the first column of the first page of the tible, and directly opposite to it is its logarithm: thus the log of 98 is 1.991226.
2. When the given number is between 100 and 1000.- Find in some of the following pages, the given number in the first column under N, and opposite to it in the next columi), marked 0 at the top, is the decimal part of the logarithm required, before which pnt an index less than the number of figures; thus log. of 448 is 2.651278.
3. When the given number consists of 4 places.--Find, as before, the first three figures of the given number in some of the colunins on the left hand, and the fourth figure at the top or bottom of the page; then die rectly under the fourth figure, and in a straight line with the three first figures on the left, will be the decimal part of the logarithm sought, before which put the index 1 less than the figures; thus, the log. of 6704 is 3.760724.
4. When the given number consists of more than 4 places.-Find, as above, the logarithm of the first 4 figures, then multiply the difference between this logarithm and the next greater in the table, by the remain. ing figures of the given number, and from the right hand of the prodact cut off as many figures as you multiplied lov, adding the remaining figures to the logarithm already taken from the table; then will the sum be the decimal part of the logarithin sought, before which write its proper index, as before, Example.-Required the logarithm of 365154.6? Here logarithm of 365100
5.562412 And tab. diff. 119 X 54.6 = 64.976 or nearly = 65 Hence, logarithm of 363154.6
= 5.562-477 When the given number is a vulgar fraction.--Subtract the logarithm of the denominator from the logarithm of the numerator, and the remainiler will be the logarithm of the given fraction,
Ex. 1. Find the log. of to Er. 2. Find the log. of 7 or
To find the natural number answering to any given logarithm.-Look in the different colun!ns for the decimal part of the given logarithm ; but if you cannot fiod it exactly, take the next less tabular logarithm, and in a line with the log. found in the column on the left marked N, you have three figures of the number sought, and at the top of the column in which the log. is, you have one figure more, which anuex to the other three ; then subtract the tabular log. from the given log. and annex two cyphers to the remainder, divide the result by the tabular difference, and annex the quotient to the four figures already fuuod. In placing the decimal point, it must be remembered, that the number of integer places in the natural number sought, is one more than the index of the given logarithm. Thus an index of 1 requires two whole numbers, and of 2 three whole numbers.
Ex. Find the natural number answering to the logarithm 3.562477.
6500 + 119
54 Hence, the natural number required
LOGARITHMICAL ARITHMETIC. Ist. The sum of the logarithms of two or more numbers is equal to the logarithms of the product of these numbers;
2d. The difference of the logarithms of two numbers is equal to the logarithm of the quotient of these numbers;
3d. The logarithm of any power of a number is equal to the logarithm of the root multiplied by the index of the power, and
4th. The logarithm of any root of a number is equal to the logarithm of that number divided by the index of the root. On these four properties the following rules are founded. TABLES OF LOGARITHMIC SINES, TANGENTS,
SECANTS, &c. The Logarithmic Sines, Tangents, Secants, &c. are the logarithms
of the natural numbers which express the measure of the sine, tangent,or secant of the corresponding arc, the radius being 10,000,000,000, the logarithm of which is 10.
Prob. 1.—To find the logarithmic sine, tengent, secant, &c. of any number of degrees and minutes.
Rule. When the number of degrees is less than 45o, find them at the top of some page, and opposite to the minutes on the left-band, under the words sine, tangent, or secant, respectively, you have the logarithm required.
2. When the number of degrees is above 45°, and less than 90°, find them at the bottom of the page, then opposite to the minutes in the righthand columns, and above the words sine, tangent, or secant, respectively, you have the logarithm required.
3. When the number of degrees is between 90° and 180°, take their supplement to 180° ; when between 180° and 270°, diminish them by 180° ; when between 270° and 360°, take their complement to 360° ; and find the logarithm of the remainder as before. Otherwise, for the log. sine or tangent of an arc between 90° and 180°, or between 270o and 360°, take out the log. co-sine, or log. co-tangent of the excess of the arc above 90° or 270°; for log. co-sine, or log. co.tangent, of an arc above 90° or 270°, take out the log. sine, or log. tangent, of the excess of the arc above 90° or 270o. But for the log. sine and log. tangent, &c. of an arc between 180° and 270o, take the log. sine and log. tangent, &c. of the excess of the are above 180.
EXAMPLES. 1. Of 250.45' the sine is 9.687935, and the tangent 9.683356. 2. Of 30°.19' the secant is 10.063864, and the co-sine 9. 936136. 3. Of 65o.Sý the sine is 9.959310, and the co-tangent 9.657028. 4. Of 749.20' the co-secant is 10.0164 12, and the tangent 10.552130. 5. Of 129o.10' the sine is 9.884477, that is, either the sine of 50°50', or
the co-sine of 390.10' 6. Of 3000.30' the tangent is 10.229832, that is, either the tangent of 590.30',
or the co-tangent of 300,30'. 7. Of 220°.18' the sine is 9.810763, that is, the sine of 40°18'.
Prob. 2.– To find the logarithmic sine, tangent, or secant, &c. of any number of degrees, minutes, and seconds.
Rule. Find, as before, the logarithm for the given degrees and minutes; then, multiply the tabular difference taken from the column marked D, by the given number of seconds; from the product cut off two decimal places, and add to the logarithm already found, the figures which remain, then will the sum be the logarithm for the degrees, minutes, and seconds, required.
Ex. Find the logarithmic sine of 27°.186.42'.
Hence, the logarithmic sine of 270.18.42" = 9.661652 When the logarithmic sine, tangent, or secant, of an angle under 3° is wanted, it may be found by the following
Rule. From the common logarithm of the number of seconds in the given angle, subtract the logarithm of the seconds in the degrees and minutes next lower, add the remainder to the logarithmic sine, tangent, or secant, of the degrees and minutes; the sum will be the logarithmic sine, tangent, or sccant, required.