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*+x.x*--************+&c. But for the same reasons as above, r must be indefinitely small, and therefore may be rejected from the factors 1-1,*9-2, 1-3, &c. Consequently, taking 1 from each side of the above equa

rr2 rr rxt rr tion, a=-r24

&c. But-rad and therefore, dividing the left hand side of the equation

72 73.1 by and the other byếr we have ap=r+++;

2

3

5

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2

3

5

++ &c. =the hyperbolic logarithm of M. 11. As, by the last article, the hyperbolic logarithm of N x2

xs

r7 or 1+ris - +

+

-&c. and as

x2 x2 the hyperbolic logarithm of Mor 1-s is x+++ 25 25 27

+ + + &c. the hyperbolic logarithm of NXM,

6

7

+

6

1

1+1

is equal to the sum of these two series, that is,

or

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3

5

7

5

n

2.33 2.25 2.x? equal to 2x + + +

+ &c. This series converges faster than either of the preceding, and its valae may

XS x? be expressed thus : 2x4+++ -+

7

+ &c.) inti

en+2 12. The logarithm of =2x logarithm of

2n+it, 2n + 1 the logarithm of

For as the addition of lo

2n +11 garithms answers to the multiplication of the numbers to which they belong, the logarithm of the square of any num. ber, is the logarithm of the number multiplied by 2. Hence

2n+2 the logarithm of is 2 x logarithm of But 2n + 1

2n+1 2n+212 2n+n 2n+212

4n* +8n+4 X 2n +11 2n+1)-1 2n+11'-1

4n2+4n na +2n+1 n+1xn+1_n+1

nun nxn+

2n +22

2

2

n

2n+ 1g and

From the preceding articles hyperbolic logarithms may be calculated, as in the following examples.

Example 1. Required the hyperbolic logarithm of 2. Put 1+1

2n + 2 4 2n +11 = 2, and then n=l,

2n+1-1 9

In order to proceed by the series in article 11, let 8 Ito 4

and then I Consequently,
3

x =0.14285714286
r3

=0.00097181730
3

It

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=0.00001189980

=0.00000017347

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[ocr errors]

=0.00000000004.

11
Sum of the above terms 0.14384103622

2
1+con+2 4
Log. of

or

or -0.28768207244 17 an+ 1 3 The double of which is 0.57535414488, and answers te the first part of the expression in article 12.

1+x

9 Secondly, let

and then 8 + 88=9-9x, and 8

1 z now is equal to

17. Consequently,

1=0.05882352941
203

=0.00006784721

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& log i lor la

=0.00000014086

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=0.00000000035

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4

3

which answers to the second part of the expression in article 12. Consequently the hyperbolic logarithm of the number 2 is 0.57536414488 +0.1177880356=0.69514718054.

The hyperbolic logarithm of 2 being thus found, that of 4, 8, 16, and all the other powers of 2, may be obtained by multiplying the logarithm of 2, by 2, 3, 4, &c. respectively, as is evident from the properties of logarithms stated in article 6. Thus, by multiplication, the hyperbolic logarithm of 4 =1.38629436103 of 8=2.07944154162 &c.

From the above the logarithm of 3 may easily be obtained. For 4= 3=4*1 =3; and therefore as the logarithm of

3 was determined above, and also the logarithm of 4, Froin the logarithm of 4, viz. -- 1.98629436108,

4 Subtract the logarithm of viz. 0.28768207244,

s' And the logarithm of 3 is 1.09861908864. Having found the logarithms of 2 and 3, we can find, by addition only, the logarithms of all the powers of 2 and 3, and also the logarithms of all the numbers which can be produced by multiplication from 2 and 3. Thus,

To the logarithm of 3, viz., 1.09861223564
Add the logarithun of 2, viz. 0.693 14718054

And the sum is the logarithm of 6 - 1.79175946918. . To this last found add the logarithm of 2, and the sum 2.48490664972 is the logarithm of 12.

The hyperbolic logarithms of other prime numbers may be more readily calculated by attending to the following article.

13. Let a, b, c, be three numbers in arithmetical progression, whose common difference is 1. Let b be the prime number, whose logarithm is sought, and a and c even numbers whose logarithms are known, or easily obtained from others already computed. Then, a being the least of the three, and the common difference being 1, a -b-l, and c=b+1. Consequently axc=b-1*b+1

2

ac+1 b-'), and actlb?; and therefore

This is a general expression for the fraction which it will be

ac

ac

1+1 proper to put 1

that the series expressing the hyper

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[ocr errors]

ac

bolic logarithm may converge quickly. For as act1 ac tacx=act1- acx — , and therefore 2 acr

1 +1=1, and r =

2act Example 2. Required the hyperbolic logarithm of 5.

1 Here a=4, c=6, and r=

Consequently, 2ac+1

49
=0.0204081632

[ocr errors]

=0.000002833%
3
ars

=0.0000000007
5

Sum of the above terms

0.0204109971

1+1 25 Log. of or

0.0408219942 1r 24 25 But x8x3=25, and the addition of Logarithms an

24 swers to the multiplication of the natural numbers to which they belong. Consequently,

25
To the log. of

0.0408219942
24
Add the log, of 8

2.0794415422
And also the log. of 3

1.0986122890

And the sum is the log. of 25 - 3.2188758254 The half of this, viz. 1.6094379127, is the hyperbolic logarithm of 5; for 5 x 5=25. Example 3. Required the hyperbolic logarithm of 7.

1 1

ac+1

1x Here a=6,c=8, and x =

and

+1 Luc+ 97 Consequently, 48

oc

49

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The sum is the log. of 49 - - 3.89182029798

49 For

48

X6x8=49. Consequently the half of this, viz. 1.94591014899, is the hyperbolic logarithm of 7; for 7 X7=49.

If the reader perfectly understand the investigations and examples already given, he will find no difficulty in calculating the hyperbolic logarithms of higher prime numbers. It will only be necessary for him, in order to guard against any embarrassment, to compute them as they advance in succession above those already mentioned. Thus, after what has been done, it would be proper, first of all, to calculate the hyperbolic logarithm of 11, then that of 13, &c.

Proceeding according to the method already explained, it will be found that The hyperbolic logarithm of 11 is 2.397895273016

of 13 is 2.564999357538
of 17 is 2.833213344878

of 19 is 2.944438979941 Logarithms were invented by Lord Neper, Baron of Merchiston, in Scotland. In the year 1614, he published at Edinburgh a small quarto, containing tables of them, of the hyperbolic kind, and an account of their construction and use. The discovery afforded the highest pleasure to mathematicians, as they were fully sensible of the very great utility of logarithms; but it was soon suggested by Mr. Briggs, afterwards Savilian Professor of Geometry in Oxford, that another kind of logarithms would be more

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