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85. The great practical utility of logarithms has induced mathematicians to expend much labor in the construction and correction of tables, in which are given the logarithms of all numbers from I to 10,000, to 100,000, and in some to 1,000,000; and this not only to 5 places of decimals, but to 7, 10, and even 14 places. Partial tables have been constructed to 20 places, and the logarithms of all the prime numbers below 1100 have been calculated to 61 places. This extreme accuracy is of course unnecessary in practice. Indeed seven places are considered sufficient for nearly every practical purpose in astronomy and navigation.*

86. We have already seen that the construction of logarithms requires the solution of the equation


in which a is the base of a system, b any number, and a the logarithm of that number. The value of x can be found exactly only in a few cases. When the base of the system is 10, for example, the logarithms of the integral powers of 10 are 1, 2, 3, 4, &c., but the logarithms of all other numbers are incommensurable; that is, they cannot be expressed exactly by any fraction or by any number of

* LALANDE, in the preface to his pocket volume of logarithmic tables, (edition of 1805,) makes the following declaration : “I have calculated some hundreds of eclipses, and I have rarely used any other tables than those which I now publish ; for my experience proves, that observations are very rarely so accurate as to require any others.” Great accuracy, however, is now attained in observation, and seven places are very generally employed in astronomical calculations. In nautical calculations more than five places are seldom if ever used.

decimals.* We may, however, approximate to their values with any required degree of accuracy. The most convenient method of approximation is by series, which we proceed to explain.

We shall make use of the following principle.

87. Any finite number whatever may be represented by a fraction whose numerator and denominator are both indefinitely small.

Thus, the number 3 may be represented by

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in short, by any fraction whose numerator is three times its denominator; consequently, by a fraction whose numerator and denominator are indefinitely small, provided their ratio or quotient is 3.

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In general, x being any number, we may represent it by , in

1 which is x times whatever value is assigned to N. If N


1 is indefinitely great,


are indefinitely small, but their quo

N N tient is still x. We conclude, then, that if h and k are two indefin. itely small quantities whose ratio is x, we may always represent a by putting





* A rigid demonstration of this may be found in Bourdon, Algébre, p. 342. Paris, 1837.

† The algebraic symbol expressing an indefinitely small quantity is 0, and the symbol expressing an indefinitely great quantity is &. The quotient obtained by dividing any finite number by a is 0; and the quotient obtained by dividing any finite number by 0 is L. The reasoning in the text might therefore be given thus: 3 being any finite number, we may

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sion, which may represent any number whatever. See Davies' Bourdon, p. 101, et seq.


88. The method of solving the exponential equation ax=b, given in Chapter V, does not afford a general expression for the value of x in terms of a and b, and is moreover too laborious to be ot'any use in the construction of tables of logarithms. The general solution is as follows, where we first convert a and b into binomials in order to apply the binomial theorem.

h =1n

ki small, according to Art. 87. Then our equation becomes



(1+p)*=1+n. Involving this to the power k, we have, by Art. 8,

Expanding both members by formula (10) of Art. 33, we have

h(-1) h(h-1)(-2)

-p8+ &c.=1+kn





-n3+ &c.


Canceling the 1 from both members, and dividing by k, we have h h h-1 h (h-1)(-2)

k-1 p+

p3 + &c. =nt n? k 2 k 2.3


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Now h–1 differs from -1 only by the indefinitely small quantity h; we may therefore use

-1 instead of h-1, without any appre

ciable error.

For the same reason we may omit h from all the factors h-2, 2-3, &c., and k from all the factors k-1, k-2, k-3, &c. Equation (28) then becomes -1 (-1)(-2)

-1 -1( + +


n3+ &c. 2.3


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From this equation we now find the value of x by dividing by the compound factor p-p+jps- &c., which gives n-11° + n-1n*+ins &c.

(30) p-ip' + {p3 - piins- &c. in which the second member is a known quantity, since n and p are known, when a and b are given, from the equations

a=1+p, b=l+n; whence pra—1, n=6–1. These values substituted in (30) give x = log. 6=(-1)={(6—1)+}(6—1)—}(6—1)++ &c.

(31) (a—1)-{(a^1)*+ }(a—1)8—1a-1)++ &c. which is the general solution of the equation a*=b, and gives the value of the logarithm of any number b in the system whose base is a.

89. In using this formula to calculate a table of logarithms, we first find the value of the denominator, which is constant, since it involves only the base of the system. Represent this denominator by A, and its reciprocal by M; that is, let

A=(a–1)-1(a−1)2+ (0-1):— (a1)++ &c. (32) 1

1 and Me A (a-1)-1(a-1) + (a-18-1a-1)*+&c.

(33) Formula (31) will then become

log. b=M[1—1—1(6—1)*+}(6-1)-1(6-1)++ &c.] (34) This constant factor M, whose value depends upon the base of the system, is called the MODULUS.


90. This is the inverse of the preceding problem, and the solution of it must give the number in terms of its logarithm and of the base of the system (or of the modulus, which depends upon the base.) If the given logarithm were integral, the number to which it corresponded might be found by involving the base to the power indicated by the logarithm ; but in all other cases evolution would also be necessary, as a single example will show. The number whose logarithm is .3 in the system whose base is a, is qu3 or q10, or Var; hence the number will be found by involving the base to the third power and then extracting the tenth root. In the following solution, however, we obtain an expression for the required number, in which the given logarithm does not enter as an exponent. 91. Resume the fundamental equation

ax=b, or b=a“,

h and as before, let a=1+p, x= (h and k being indefinitely small.) Then we have

b=(1+p), whence b=(1+p)". Now, by the binomial theorem, we have (1+p)'=1+hp+ h(h—122+h(h—1)(h—2) (35)


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'p3 + &c.

Neglecting h in all the factors h1, h2, &c. this equation becomes, as in Art. 88,

(1+p)=1+h(p— įp?+ } p3— &c.)

=l+h[a-1-a-1)+1(a--1)3— &c.]

=1+hA, (Art. 89) that is,

b=1+hA, whence




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