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363

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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

b

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

+

343

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 =

d

i Sa =)

band again, ass:b::1:1+*; * =

d (25=-=) : : Therefore substituting

for *, we thall have the Logarithms of those Ratios; viz.

d

d2 d3 d4
+
+
+

&c. and

33 I

d4 +

&c.

333
The Sum of which two Logarithms, viz.

d
d 3

d?
X 2 X

+ +

GC. is 35

the

I
Sa

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I

n

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+

n

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sod

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

+ +

+

&c. whose

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,

and b

any

Number at Pleasure ; then d bwhich allume = e, and then b=

ite bts

ss-dd = 2*

S

53

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555

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S

I

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=

[ocr errors]

I

and because = e, therefore have we for

THEOREM I.

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n

The Log.ofb=

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,

EXAMPLE.

i.

Here (6=)?He=2;

=2; therefore I te= (1-c *2 =) 22e; and 3e = (2-1=) 1; whence e = }, and

Ie

The The OPERATION stands thus:

[ocr errors]

e3

[ocr errors]
[ocr errors]

=,33333333

-,33333333
3703704

1234568
411523

82305
45725

6532
5081

565
565

59
63

5
234657359

2
Whence Neper's Logarithm of 2 is ,69314718
But ,69314718, multiplied by 3, will give 2,07944154
for the Logarithm of 8, inasmuch as 8 is the Cube or
third Power of 2; and the Logarithm of 8 + Log. of
1 is equal to the Logarithm of 10, because 8 X1 =10;
wherefore to find the Logarithm of I we have b=
Ite

=I=; whence es, and ee = x's

The OPERATION stands thus :

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e=UTI

=IIIIIIII
e3 137174

45725
1693

339'
21
1e7

3
,11157178

2
Whence Neper's Logarithm of 1.is ,22314356
To which add the Logarithm of 8, 2,07944154
The Sum, viz.

2,30258510
is Neper's Logarithm of 10. But if the Logarithm of
10 be made 1,000000, &c. as it is for Conveniency

2302585
done in most of the Tables extant, then

n

1,004, &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 sufficient, the Index
x would have been 2,30258, 50929, 94045, 68401,

799142

n

79914, &c. and its Reciprocal, viz. = 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. :::

&c. :: 2=,868588 9638, &c. which put=m, and then the Logarithm of Ite

me? b

me +

+

+ +
3 5 7 9

n

n

me3

mes

mes

&C.

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Let it be required to find Briggs's Logarithm of 2.
Here b= * = 2 ::e=;, and ee = šo

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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= 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

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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

Ie

The OPERATION is as follows:

[ocr errors]

me

me

[ocr errors]

mes mes me? mes

me?

1588

50

men mell

II me

=,868588963
-,173717792

=,173717792 6948712

2316237 277948

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.

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