bers, is 1 +*+**+xxx+xxxx, &c. and the celebrated binomial Theorem invented by Sir Isaac Newton for x 2n -+ n 2n -3 in this Cafe rather I X X &c.for 3 4 5 being an Infinitesimal, is rejected; whence the infinite x2 x3 Root of 1 X*" I + + &c. and the 31 4n x2 x3 Excess thereof above Unity, viz.. 3n 4n &c. is the Argument of the first of the mean Proportionals between Unity and. 1 **, which therefore will be as the Logarithm of the Ratio of 1 to 1 +*, or as the Logarithm of 1 +*. But as 1+*- 1 is a Ratiuncula, it must be multiplied by 10000, &c. infinitely, which will reduce it to Terms fit for Practice, makeing the Logarithm of the Ratio of 1 to it * = + &c. whence if the 3 4 Index n be taken 1000, &c. as a Neper's Form, the x4 Logarithms will be fimply * -- + + 3 4 5 &c. But as n may be taken at Pleasure, the several Scales of Logarithms to such Indices will be as 1000, &c. reciprocally as their Indices. Again, if the Logarithm of a decreasing Ratio be 1000, &c. x2 I 2 2 or n n 2n sought, the infinite Root of Iow* = 1-* will be found by the like Method to be i 3 **, &c. which fubtract from Unity, and the Decre4n ment of the first of the infinite Number of Proportionals will appear to be &c. 3 4 which 1 &c. or 262 which expresses the Logarithm of the Ratio of 1 to 1-X, or the Logarithm of 1-4, according to Neper's Form, if the Index n be put =10000, &c. as before. Aid to find the Logarithm of the Ratio of any two Terms, a the leffer, and b the greater, it will be as a : 1 b (b-a ::1:1+ *; whence 1+x= -; and x = the Difference divided by the leffer Term when 'tis an b. increasing Ratio, and when 'tis decreasing. b Wherefore, putting d=Difference between the two Terms. a and b, the Logarithms of the fame Ratio may be doubly exprefied, and accordingly is either d d3 04 Х + * d d th d -X + + &c. both producing 3 + b the fame Thing. But if the Ratio of a to b be supposed to be divided into two Parts, viz. into the Ratio of a to the arithmetical Mean between the two Terms, and the Ratio of the said arithmetical Mean to the other Term b, then will the Sum of the Logarithms of those two Ratios be the Logarithm of the Ratio of a to b. Wherefore subfticuting is for Latib, and it will be, {s: d @::1:1-*; whence * = es and again, ass:b::1:1 + x ; x = 음 26 d : Therefore substituting for Thall have the Logarithms of those Ratios; viz. d c. and + + + 262 363 3 464 I 5,24 =) S (20= =) *, we d3 2x + ss-dd 13 the Logarithm of the Ratio of a to b, whose Difference is d, and Sums s; which Series, without the Index n, is, by-the-bye, the Fluent of the Fluxion of the std Logarithm of assuming d, to be the Aowing sed Quantity, for the Fluxion of the Logarithm of std 2sd d dad d41 død is + + s? +, &c. whore d a3 d? Fluent 2 x -t + &c. is Neper's 757' and the same as above, abating sod the Index n. This Series, either Way obtained, converges twice as swift as the former, and consequently is more proper for the Practice of making Logarithms : Thus put a=1, and b any Number at Pleasure ; then d which assume and then b = Ite. 6+1 d and because =l, therefore have we for 353 555 Logarithm of std, S THE OREM I. 2 n te The Log.ofb= Xet es + šos + . To illustrate this Theorem: Let it be required to find the Logarithm of 2 true to 7 Places. Note, That the Index must be assumed of a Fi gure or two more than the intended Logarithm is to have. EXAMPLE. Mere (b='+e=2; therefore i te=(1x2 =) 2--21; and 3e= (2-1=) 1; whence e = 1, and le= The Whence Neper's Logarithm of 2 is ,6931471 ==; whence e = 1, and ee = The OPERATION stands thus : e IIIIIIIIرد = =,I1IITTI 137 174 1693 21 45725 3 Whence Neper's Logarithm of 1 is ,22314356 2,30258510 2302585 1,000,&c. Whence n=2302585, &c. is the Index 1 Here b == 79914, &c. and its Reciprocal, yz. = 0,43429, 44819, 03251, 82765, 11289, &c. which, by the Way, is the Subtangent of the Curve expressing Briggs's Logarithms; from the Double of which the faid Logarithms may be had directly. For, because I = 0,4342944, &c. *** =,868588 9638, &c. which put=m, and then the Logarithm of b & C. 3 5 7 9 EXAMPLE. Let it be required to find Briggs's Logarithm of 2. = 2:::e= and ee = 5. 2 ņ n Ite 3 me mes me? mel me + + + Let it be required to find Briggs's Logarithm of 3. Now because the Logarithm of 3 is equal to the Logarithm of 2 plus the Logarithm of 1 (for 2 X= 3), therefore find the Logarithm'of 1, and add it to the Logarithm of 2 already found, the Sum will be the Logarithm of 3, which is better than finding the Logarithm of 3 by the Theorem directly, inasmuch as it will not converge so fast as the Logarithm of 1; for the smaller the Fraction represented by e, which is deduced |