LOGARITHMS. 



Term of the series. 

 0200000000000000 = tirst 



6666666666 = second 

 400000 = third 

 28 = fourth 



Sum = 0.02u0006667066694 = area b D. 



Term of the series. 

 0.0001000000000000 = rirst 



50000000 = second 

 3333 = third 



0.0001000050003333 = area Ad AD. 



Half the aggregate 0.0100503358535014 

 = Ad, and half the remainder, viz. 

 0.0099503308531681 == A D. 



And so putting A B = A b = y-i-- or 



C B = 1.001, and Ct> =* 0.999, there will 

 be obtained A (/ = 0.00100050003335835, 

 and A D = 0.00099950013330835. 



After the same manner, if A B = A 6, 

 be = 0.2, or 0.02, or 0.002, these areas 

 will arise. 



A d = 0.2231435513142097, and 

 AD == 0.1823215576939546, or 

 A d = 0.0202027073175194, and 

 AD = 0.1098026272961797, or 

 A d ; = 0.002002, and A D = 0.001. 

 From these areas, thus found, others 

 may be easily had from addition and sub- 



12 12 

 traction only. For since --{ ^ = 2, 



the sum of the arcs belonging to the ra- 



1.2 1.2 

 tios ~ and ^ (that is, insisting upon 



the parts of the absciss 1.2, 0.2 ; and 1.8, 

 0.9), vis. 



added thus, 



Sum = 0.28768, &c. 



Total = 69314, &c. == the area 

 of A F H G, when C G is 2. Also since 



H + 2 = 3, the sum 1.0986122, &c. of 



1 2 

 the areas belonging to and 2, will be 



U.o 



the area of A F G H, when C G = 3. 



2x2 

 Again, since ^- = 5, and 2 X 5 = 



10 ; by adding A d 0.2231, &c. A D = 

 0.1823, &c. and A d == 0.1053, &c. toge- 

 ther, their sum is 0.5108, &c. and this 

 added to 1.0986, &c. the a^ea of A F G H, 

 VOL. IV. 



when CG = 3. You will have 

 1.6093379124341004 = A F G H, when 

 C G is 5 ; and adding that of 2 to this, 

 gives 2.302585092994J457 = A F G 11, 

 when C G is equal to 10 : and since 10 X 

 10 = 1Q ; and 10 x 100 = lOGu ; and 

 V5x 10 X 0.9J~=- 7, and 10 x 1.1 = 

 1000 X 1.091 



11, and 

 1000 X 0.998 



7 X 11 



13, and 



499 ; it is plain that the 



area A F G H may be found by the com- 

 position of the areas found before, when 

 C G = 100, 1000, or any other of the 

 numbers above-mentioned ; and all these 

 areas are the hyperbolic logarithms of 

 those several numbers. 



Having thus obtained the hyperbolic 

 logarithms of the numbers 10, 0.98, 0.99, 

 1.01, 1.02; if the logarithms of the four 

 last of them be divided by the hyperbo- 

 lic logarithm 2.3025850, &c. of 10, and 

 the index 2 be added ; or, which is the 

 same thing, if it be multiplied by its re- 

 ciprocal 0.4342944819032518, the value of 

 the subtangent of the logarithmic curve, 

 to which Brigg's logarithms are adapted, 

 we shall have the true tabular logarithms 

 of 98, 99, 100, 101, 102. These are to 

 be interpolated by ten intervals, and then 

 we shall have the logarithms of all the 

 numbers between 980 and 1020 ; and all 

 between 980 and 1000, being again inter- 

 polated by ten intervals, the table will be 

 as it were constructed. Then from these 

 we are to get the logarithms of all the 

 prime numbers, and their multiples less 

 than 100, which may be done by addition 



10 

 and subtraction only : for 



V8 X 9963 

 984 



; 



10 



= 5; 



986 

 984 



9911_ 

 Tl X~l 

 = 61; 



= 73; 

 = 89; 



'"f^,-"" 



r = -^ 6 - 



29;^ = 



^98 



2 



102 __ 

 6 = 



= 23; 



989 

 IT 



9971 



T ' 13 X 13 



9949 ' _ 994 



3X49 

 9954 



71*18 

 9894 



6 X17 



999 



1 W 

 987 

 21 



= 59; 



= 47; 

 9882 



2X81 

 9928 



_.^L 8 3.^i 



' 12 ~ '7X16 

 97 ; and thus hav- 



ing the logarithms of all the numbers less 



