log. 13=

SGS5S

1

1 3-log. 7-log. 11+0.96858896 +

+&c.

2001' 3(2001) 3
2001 0.86858896
43407=sum of series.

log. 7=0.84509804
log. 11=1.04139269

1.88649073

3.00043407
1.88649073

1.11394334=log. 13.

We might proceed in this way until we should have calculated the logarithms of all the prime members within the limits of the tables.

(226.) We have already (Art. 220) said that the base a of the Napierian system of logarithms satisfies the following equation : (a-1)-3(a−1)2 +}(a-1)3 – (a1)*+&c.=l. (1)

1 From example 3, page 231, we see that if we have

(y-1)2 , (y-1)3 , (y-1)4 (y-1)

+

+ &c.=x, 2

(2)

4 then will

23

X4 y=l+x+ot +

+ &c. 2 2.3' 2.3.4

(3) Equation (2) will agree with (1), when y=a; and x=1.

= Making these changes in (3), we find

+

3

x2

1 1 a=1+1+-+

2' 2.3

+

1

1
+ + &c.
2.3.4 ' 2.3.4.5

(4)

This series may be summed as follows:

2.71828180=base of Napierian logarithms.

This value when extended to 35 decimals is found to be e=2.71828182845904523536028747135266249.

EXPONENTIAL THEOREM.

(227.) The above theorem makes known the law of the development of a* according to the ascending powers of x.

To determine this law, we will assume

a=1+Ax+Bx2+Cx3+Dx1+&c.,

both members of which become 1 when x=0.

(1)

Changing x into y, in (1), and we have

a2=1+Ay+By2+Cy3+Dy1+&c.

(2)

Subtracting (2) from (1), and actually dividing the righthand member by x-y, we obtain

ax-a3

~=A+B(x+y)+C(x2+xy+y2) x−y +D(x3+x2y+xy2+y3)+&c.

Writing x-y for x in (1), and it becomes

Ꮖ

(3)

Transposing the 1, and multiplying by a”, we get a"(a2—'—1)=a" { A(x−y)+B(x−y)2+C(x−y)3+&c.} (5) Divided (5) by x-y, after replacing its left-hand member by its equivalent value a-a", we find

ax-ay

x-y

=a3 { A+B(x−y)+C(x−y)2+D(x−y)3+&c.} (6)

Equating the right-hand members of (3) and (6), we have A+B(x+y)+C(x2+xy+y2)

( us + wey + aye + y23) + &c. }

3

2

=a3 { A+B(x−y)+C(x−y)2+D(x−y)3+&c. }

This is true for all values of x and y.

When y=x, it becomes

A+2Bx+3Cx2+4Dx3 +&c.=a*. A.

For a, substituting its value equation (1), we find

(7)

(8)

(9)

Equating the coefficients of the like powers of x, Art. 183,

+

+

(10)

十

Hence, (1) becomes

A r2 A 23 A* 2*

A’x A4x4 qw=1+ Axt?

+&c. 2 2.3 2.3.4 It now remains to find the value of A.

For this purpose put 1+b=a, and we have a*=(1+5)*, which by the Binomial Theorem becomes

cb r(x-1)^2 =(x-1)(x-2) * (1+5)*=1++

+

+&c.

(11) 1 1.2

1.2.3 Performing the multiplications indicated, we find the coefficient of the first power of x to be

b 22 23 24

64
+ +&c.,

1 2 3 4
or resubstituting a-1 for b, it becomes

.—)2
(a-1)(a-1)_(a-1)

3 *
(a-1)-

+-&c.
2
3

4 Therefore

(a, 3
-1)? / (4-1)o_(a−1)*

-)
A=(a–1)

+&c. (12) 2 3

4 If in formula (C'), we put a-1 for n, we shall find

(a−1)2 (a−1)2 (a–1)4 Nlog. a=(a-1)- +

+&c.
2
3

4
Hence,
Å=N log. a.

(13) This value of A substituted in (10), gives

(N log. a)2.x2 (Nlog.a)3. <3 a*=1+N log. a.x+

+ (A) 2

2.3 When a=e=2.7182818&c., then

Nlog. a=N log. e=1, and (A) becomes

X 3
€*=1+x+
e

+
2 2.3' 2.3.4

(B)

+

+

3

x4 +

+ &c.

APPLICATION OF LOGARITHMS.

(228.) By the aid of a table of logarithms, we can easily perform the following operations:

1. Find the value of

3.75X1.06
365

by logarithms.

Recollecting (Art. 216) that the logarithm of the product of several factors is equal to the sum of their respective logarithms; and (Art. 218) the logarithm of the quotient of one quantity divided by another is equal to the logarithm of the dividend diminished by the logarithm of the divisor, we find for the logarithm of our expression

Therefore, the above expression is nearly equal to 0.01089.

1

2. Find the 11th root of 11, that is, the value of 11.

Taking the logarithm of this expression, we find

log. 111 of log. 11 of 1.0413927=0.0946721 =log. 1.24357&c.

Therefore,

111=1.24357.