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which expreffes the Logarithm of the Ratio of 1 to 1-4, or the Logarithm of 1-*, according to Nepers 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 : 5
16-a ::1:1+ x;
whence Itxa ; and x = the Difference divided by the lefser Term when 'tis an
bincreasing Ratio, and when’tis decreasing
b Wherefore, putting d=Difference between the two Terms a and b, the Logarithms of the same Ratio may be doubly expressed, and accordingly is either
d dz d3 d4
d d +
& c. both producing b 262 + + 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 Logarithin of the Ratio of a to b. Wherefore substituting is for at b, and it will be, is; @::1:1-X; whence x =
i Sa =)
band again, ass:b::1:1+*; * =
d (25=-=) : : Therefore substituting
for *, we thall have the Logarithms of those Ratios; viz.
d2 d3 d4
GC. is 35
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 Logarithm of s+d, affuming d, to be the Aowing
std Quantity, for the Fluxion of the Logarithm of
sed 2s à
d dad d4d død is
to 55 57
d3 ds d? Fluent 2 x
&c. is Neper's 353
757 std Logarithm of
and the same as above, abating the Index n. This Series, either Way obtained, con verges twice as swift as the former, and consequently is more proper for the Practice of making Logarithms : Thus put al,
Number at Pleasure ; then d bwhich allume = e, and then b=
ss-dd = 2*
and because = e, therefore have we for
Xet je tes,&c. 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,
=2; therefore I te= (1-c *2 =) 22e; and 3e = (2-1=) 1; whence e = }, and
The OPERATION stands thus:
=I=; whence es, and ee = x's
The OPERATION stands thus :
1,004, &c. Whence n=2302585, &c. is the Index
79914, &c. and its Reciprocal, viz. = 0,43429,
&c. :: 2=,868588 9638, &c. which put=m, and then the Logarithm of Ite
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 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 from the Number whose Logarithm is fought, the swifter does the Series converge.
Here b = TH=;":24+2=3–34:1= }, and ec s
The OPERATION is as follows:
mes mes me? mes
55590 T1118 445 18
? Briggs's Logarithm of 11
917609r259 To which add the Logarithm of 2= ,301029993 The Sun is the Logarithm of 3= 0,477121252
Again, to find the Logarithm of 4, because 2 x2=4, there ore 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 =5, therefore from the Logarithm of 10 1,000000000 fubtract the Logarithm of 2
,301029993 There remains the Logarithm of 5 = ,698970007 And because 2*3=6; therefore
To find the Logarithm of 6, To the Logarithm of 3
9477121252 Add the Logarithm of 2
2301029993 The Sum will be the Logarithm of 6 = 2778151245 Which being known, the Logarithm of 7, the next prime Number, may be easily found by the Theorem; for because 6x2=7, therefore to the Logarithm of b add the Logarithm of %, and the Sum will be the Logarithm of 7.
E X A M P L E.
Here b =+= e = is, and ee = 105.