0.0963101798043249,, will 
area A D. 
By the same tables, these areas A D and 
A d, will be obtained also when A B -j- A 6 
are supposed to be or C B = 1.01, and 
C 6 = 0.99, if the numbers are but duly 
transferred to lower places, as 
Term of the series. 
0.0200000000000000 = first 
6666666666 — second 
400000 = third 
28 = fourth 
Sum=0. 0200006667066694 = area 6 D. 
LOGARITHMS. 
be the lesser ing Ad — 0.2231, &c. A D = 0.1823, &c. 
Term of the series. 
0.0001000000000000 = first 
50000000 = second 
3333 = third 
0.00010000.50003333 = area A d — A D. 
Half the aggregate 0.01003033.5853.9014 
= A d, and half the remainder, viz. 
0.0099503308331681 = A D. 
And so putting A B = A 5 = or 
C B = 1.001 and Cb = 0 999, there will 
be obtained Ad 0.00100030003335835, 
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 = 0.223143.5513142097, and 
AD = 0.1823215576939546, or 
A d = 0.020202707317.9194, and 
AD = 0.1098026272961797, or 
A d = 0.002002, and AD = 0.001. 
From these areas, thus found, others may 
be easily had from addition and subtraction 
only. " 
1.2 
For since ^ -4- = 2, the sura of 
0.8 o.y 
added thus 
? 0.40546, &c. 
’ U- 
0.8 
areas belonging to 
1.2 
and Ad = 0.1053, &c. together, their sum 
is 0.5108, &c. and this added to 1.0986, &c, 
the area of A F G H, when C G = 3. You 
will have 1.6093379124341004 = A F G H, 
when CG is 5; and adding that of 2 to 
this, gives 2.30258.50929940457 = AFGH, 
when C G is equal to 10 : and since 10 X 
10 = 100 ; and 10 X 100 = 1000 ; and 
5 X 10 X 0.98 = 7, and 10 X 1.1 = 11, 
12 12 
the arcs belonging to the ratios ~ and ^ 
(that is, insisting upon the parts of the ab- 
sciss 1.2, 0.8 ; and 1.2, 0.9), mz. 
5 ad z= 0.18232, &c. 
0.40j46j, &c. and ^ ^ ^ _ q 
Sum = 0.28768, &c. 
t6, &c. 
28768, &c. 
and 
1000 X 1.091 
z= 13, and 
1000 X 0.998 
7 X 11 2 
= 499 ; it is plain that the area AFGH 
may be found by the composition of the 
areas found before, when CG = 100, 1000, 
or any other of the numbers above-mention- 
ed ; and all these areas are the hyperbolic 
logarithms of those several numbers. 
Having thus obtained tlie 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 hyperbolic 
logarithm 2.3025850, &c. of 10, and the 
index 2, be added ; or, which is the same 
thing, if it be multiplied by its reciprocal 
0.4342944819032518, the value of the sub- 
tangent of the logarithmic curve, to which 
Briggs’s logarithms are adapted, we shall 
have the true tabular logarithms of 98, 99, 
100, 101, 102. These are to be interpo- 
lated by ten intervals, and then we shall 
have tlie logarithms of all the numbers be- 
tween 980 and 1020; and all between 980 
and 1000, being again interpolated by ten 
intervals, the table will be as it were con- 
structed. 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 and subtraction 
only: for 
84 X 1020 _ g . 8 X 9963 
= 3 
9945 
Total = 0.69314, &c. = the area of 
A F H G, when C G is — 2. Also since 
1 2 
4 - 2 = 3, the sum 1.0986122, &c. of the 
^ y and 2, will be the 
area of AFGH, when C G = 3. Again, 
»‘Hce = 5, and 2 X 5 = 10 ; by add- 
= 11 ; 
SSp 
1001 
= 19; 
7 X 11 
9936 
16 X 27 
999 
= 31;- = 37; 
;47; 
= 61 
= 67; 
9911 
11 X 17 
, 9949 
2 X 81 ’ 3 X 49 
9928 _ 9954 
8 X 17 “ ’ 7 X 18 
9968 _ 9894 
— 89 ; ■ 
996 
= 79; — = 83; 
12 ’ 7 X 16 ' 6 X 17 
= 97 ; and thus having the logarithms of all 
the numbers less than 100, you have nothing 
