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bers, is 1+x+x+xxx+xxxx, &c, and the celebrated binomial Theorem invented by Sir Ifaac Newton for

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n

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3 4 5

being an Infinitefimal, is rejected; whence the infinite

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x3 x4

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c. and the

n

2n 3n 4n'

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2n 3n 4.n

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&c. is the Argument of the firft of the mean Proportionals between Unity and 1+x, which therefore will be as the Logarithm of the Ratio of 1 to 1+x, or as the

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n

Logarithm of 1+x. But as 1+x-1 is a Ratiuncula, it must be multiplied by 10000, &. infinitely, which will reduce it to Terms fit for Practice, makeing the Logarithm of the Ratio of 1 to 1 + x = 1000, &c. x x2 &c. whence if the

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2 3 4 5 &c. But as n may be taken at Pleasure, the feveral scales

of Logarithms to fuch Indices will be as

reciprocally as their Indices.

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Again, if the Logarithm of a decreafing Ratio be

fought, the infinite Root of 1x = 1 -x will be

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&c. which fubtract from Unity, and the Decrement of the first of the infinite Number of Proportio

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which expreffes the Logarithm of the Ratio of 1 to 1-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

::1:1 + x; whence 1+x=

b

a

; and x = =(b=a, or)

a

the Difference divided by the leffer Term when 'tis an

increafing Ratio, and when 'tis decreafing.

b- a
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

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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 thofe two Ratios be the Logarithm of the Ratio of a to b. Wherefore fubftitutings for a+b, and it will be,s: a:: I: I-, -x; whence x=

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and again, ass: b : : 1 : 1 + x ; x =

d

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2 b = = ) 2 : Therefore fubftituting

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fhall have the Logarithms of those Ratios; viz.

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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 s+d Logarithm of affuming d, to be the flowing s-d

Quantity, for the Fluxion of the Logarithm of

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s+d,

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+,&c. whofe

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Logarithm of 5+ and the fame as above, abating

s-d

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 which affume =e, and then b = 1+;

d

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To illuftrate this Theorem: Let it be required to find the Logarithm of 2 true to 7 Places.

Note, That the Index must be affumed of a Figure or two more than the intended Logarithm is to have.

EXAM P L E.

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(b) =2; therefore += (1—ex2 =)

I-e

2-2e; and 3e (2—1=) 1; whence e = 1⁄2, and

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Whence Neper's Logarithm of 2 is

565
59
5

,34657359

2

,69314718

But,69314718, multiplied by 3, will give 2,07944154
for the Logarithm of 8, inafmuch as 8 is the Cube or
third Power of 2; and the Logarithm of 8+ Log. of
Iis
14 is equal to the Logarithm of 10, becaufe 8 x1=10;
wherefore to find the Logarithm of 14 we have b=
==; whence e = , and ee

Ite

Ie

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I

45725

339

=

3

,11157178

,22314356

2,07944154

2,30258510

Whence Neper's Logarithm of 14 is
To which add the Logarithm of 8,

The Sum, viz.

is Neper's Logarithm of 10. But if the Logarithm of
To be made 1,000000, &c. as it is for Conveniency

done in most of the Tables extant, then

2302585

n

1,000, &c. Whence n=2302585, &c. is the Index
for Briggs's Scale of Logarithms; and, if the above
Work had been carried on to Places fufficient, the Index
would have been 2,30258, 50929, 94045, 68401,

79914, &c. and its Reciprocal, víz. 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.

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For, because 0,4342944, &c.·.· 2,868588

η

=

n

9638, &c. which put=m, and then the Logarithm of

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+ +
5

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EXAMPLE.

Let it be required to find Briggs's Logarithm of 2.

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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 1 3), therefore find the Logarithm of I, 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 fo faft as the Logarithm of 1; for the fmaller the Fraction reprefented by e, which is

deduced

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