which expreffes the Logarithm of the Ratio of 1 to J-x, or the Logarithm of 1-x, according to Neper's Form, if the Index n be put =10000, &c. as before. And to find the Logarithm of the Ratio of any two Terms, a the leffer, and b the greater, it will be as a : b b ::1:1 + x; whence 1+x=. ; and x = a or the Difference divided by the leffer Term when 'tis an ba increafing Ratio, and when 'tis decreafing. 1 b Wherefore, putting d-Difference between the two Terms a and b, the Logarithms of the fame Ratio may be doubly expreffed, and accordingly is either 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 faid 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 fubftitutings for a+b, and it will be, a: II -x; whence x = Sa •2a S : Therefore fubftituting d s: for x, we fhall have the Logarithms of those Ratios; viz. the Logarithm of the Ratio of a to b, whofe Difference is d, and Sums s; which Series, without the Index n, is, by-the-bye, the Fluent of the Fluxion of the affuming d, to be the flowing Logarithm of Logarithm of 5+, and the fame as above, abating the Index n. This Series, either Way obtained, converges twice as fwift as the former, and confequently is more proper for the Practice of making Logarithms: Thus put a 1, and b any Number at Pleasure; then d b S b+1 which affumee, and then b= d and because == To illuftrate this Theorem: Let it be required to find the Logarithm of 2 true to 7 Places. Note, That the Index muft be affumed of a Figure or two more than the intended Logarithm is to have. EXAMPLE. Here (6)1+=2; therefore 1+e= ( I-e (e x2 =) 2-2e; and 3e (2—1=) 1; whence e, and The But ,69314718, multiplied by 3, will give 2,07944154 =1; whence e, and ee = }}• 2,30258510 is Neper's Logarithm of 10. But if the Logarithm of done in most of the Tables extant, then 2302585 n 1,000, &c. Whence n=2302585, &c. is the Index 79914, &c. and its Reciprocal, viz.= 0,43429, n 44819, 03251, 82765, 11289, &c. which, by the Way, is the Subtangent of the Curve expreffing Briggs's Logarithms; from the Double of which the faid Logarithms may be had directly. For, because 0,4342944, &c. ·.· 2=,868588 n n 9638, &c. which put=m, and then the Logarithm of Let it be required to find Briggs's Logarithm of 2. 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 × 1 = 3), therefore find the Logarithm of 11, 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, inafmuch as it will not converge fo faft as the Logarithm of 1; for the smaller the Fraction represented by e, which is deduced deduced from the Number whofe Logarithm is fought, the fwifter does the Series converge. Here b and ee=23 Ite Ie =26+2=33ee= Briggs's Logarithm of 11= To which add the Logarithm of 2= 0,477121252 Again, to find the Logarithm of 4, because 2 x2=4, therefore the Logarithm of 2 added to itself, or multiplied by 2, the Product 0,602059986 is the Logarithm of 4. To find the Logarithm of 5, because 10 = 5, therefore from the Logarithm of 10 fubtract the Logarithm of 2 1,000000000 ,301029993 There remains the Logarithm of 5= ,698970007 And because 2 x3=6; therefore To the Logarithm of 3 Add the Logarithm of 2 477121252 301029993 778151245 To find the Logarithm of 6, The Sum will be the Logarithm of 6= Which being known, the Logarithm of 7, the next prime Number, may be eafily found by the Theorem; for because 6x=7, therefore to the Logarithm of 6 add the Logarithm of, and the Sum will be the Logarithm of 7. |