*+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 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 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, x =0.14285714286 =0.00097181730 It =0.00001189980 =0.00000017347 =0.00000000004. 11 2 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 =0.00006784721 & log i lor la =0.00000014086 =0.00000000035 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 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 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 =0.000002833% =0.0000000007 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 0.0408219942 2.0794415422 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 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 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 |