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this case, have * (2acts) = 1** = zis:

whence,

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2

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1

**

Therefore{$(*+*+&c.)=,058891517&c.and

35 consequently hyp. log. 3: (byp.log. 2+ hyp. log. 4: + {S)= 1,098612288 &c.

2. Let the hyperbolic logarithm of 1o be required

The logarithms of 8 and 9 being given, from those of 2 and 3 (already found), a may, here, be =8,b=9 and c=10; and then x ( ) being

200-1

x ”
we shall have { S (x +*+ &c.)
161

3 5 ,006211180 &c. + ,000000079 &c. &c. = ,0062 11 259 &c.

And therefore log. 10 (2. log. 9- log. 8-5) = 2,302585092 &c.

Hitherto we have had regard to logarithms of the hyperbolic kind: but those of any other kind

may be derived from these, by, barely, multiplying by the proper multiplicator, or modulus.

Thus, in the Brigean (or common) form, where an unit is assumed for the logarithm of 10, the logarithm of any number will be found, by mul

tiplying

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tiplying the hyperbolic logarithm of the same number by the fraction ,434294481 &c. which is the proper modulus of this form.

For, since the logarithms of all forms preserve the same proportion with respect to each other, it will be, às 2,302585092 &c. the hyperbolic log. of 10 (above found) is to (H) the hyperbolic logarithm of any other number, fo is 1, the

H common logarithm of 10, to

2,302585092 &c. H®,434294481 &c. the common logarithm of the same number.

But (to avoid a tedious multiplication, which will always be required when a great degree of accuracy is insisted on the best way to find the logarithms of this form is from the series 2x + 2x)

&c. X 0,434294481 &c. which expresses, 3 5

I+x the common logarithm of (by what has been

to

( 2,302585093&c.

2x05

+

I

+

already shewn), and which, by making R = ,868588963 &c. will stand more commodiously

Rx Rx5 RX7 thus, Rx +

&c. +

3 5 7 For an example hereof, let the common logarithm of 7 be required: in which case the lo. garithms of 8 and 9 being known, from those of 2 and 3), we shall have log. 7 = 2 log. 8 -- log. 9

Rox' Rixos -S (by the Theor.), S being =Rx

to

3 S and x =

64-63)

64763 : whence (** being =

we hall have 127

161292

Rx

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&c. (=the common log of it)

I

I

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Rx (8685 &c.) =,006839283 &c.

127 Rx

Rx (161294

16129) = 1000000424 &c.

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Rx 16129) &c.

6129)

=,00000000002 &c.

,

&c.

Rxz RX5
Consequently S (RX + +

&c.) =

3 5 9006839424. &c. and 2 log. 8 – log. 9-S= ,845098040 &c. = the common logarithm of 7 required. But the fame conclusion may be brought out by fewer terms of the series, if the logarithms of the three first primes 2, 3 and 5 be sup-. pored known; because those of 48 and 50 (which are composed of them) will likewise be known; from whence the logarithm of 7 (= {log. 49 = log. 48.+log: 50+S) will come out =,845098040

4 &c. (as. before) which value will be true to 11 places of figures by taking the first term of the feries, only.

Again, let the common logarithm of the next prime number, which is II, be required. Here a may be taken = 10,6 = 11, and c = 12; but, fewer terms of the feries will suffice, if other three numbers, composed of u and the inferior primes, be taken, whereof the common difference is an unir. Thus, because 98 = 2X7 X7, 99=3* 3x11 (9x11), and 100=2x2-x5x5 (or 10x10), let there be taken a=98, b=99, and c=100; and then, by the first term of the series only, the log. of 99 will be found true to 14 places; whence that of 11 (log. 99-log. 9.) is also known.

But

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But notwithstanding all these artifices and compendiuins, à method (similar to that in page 18.) for finding the logarichins of large numbers, one from another, by addition and iubtraction, only, Itill seems wanting in the calculation of tables; I fhail, therefore, here subjoin such a method.

1. Let A, B and C denote any three numbers in arithmetical progression, not less than 10000 each, whereof the common difference is 100.

2. From twice the logarithm of B, subtract the sum of the logarithms of A and C, and let the remainder be divided by 10000.

3. Multiply the quotient by 49,5, and to the product add to part of the difference of the logarithms of A and B; then the sum will be the excess of the logarithm of À + i above chat of A.

4. From this excess let the quotient (found by Rule 2.) be continually fübtracted, that is, first from the excess itself, then from the remainder, then from the next remainder, &c. &c.

5. To the logarithm of A add the said excess, and to the fum add the first of the remainders to the last sum add the next remainder, &c. &c. then the several sums, thus arising, will exhibit the logarithms of A + 1, A + 2, A + 3, &c. respectively.

Thus, let it be proposed to find the logarithms of all the whole numbers between 17900 and 18100; those of the two extremes 17900 and 18100, and that of the mean.(18000) being given.

E

Then

{

c Jro The Nature of Logarithms. Then the loga-S A Žbeing S 4,2528530317 rithm of

B equal 4,255272505

4,257678575 we shall have

2 log. Blog. A-log. C

10000 ,00000000134 (see Rule 2.) which multiplied by 49.5, and the product added to

log. 'B - log. A

ICO gives ,00002426107 for the excess of the logarithm of A +1 above that of A (by Rule 3.) From whence the work, being continued according to Rule

4
and

5. will stand as follows: ,000024 26107 excess. 4,25|2853031 log: 17900 134

2426107 excess. 287729207 log. 17901

2425973 25839 20 rein. 290155180 log. 17902 134

2425839 25705 3d rem.

292581019 log. 17903 134

2425705 25571 4th rem.

295006724 log. 17904 134

2425571 254.37 5th rem.

297432295 log. 17905 134

2425437 25303

299857732 log. 17906 134

2425303 25169threm, 302283035 log. 17907 134

2425169 25035

304708204 log. 17908 134

2425035
24901 gb rem. 307133239 log. 17909
134

2424901
24-67 rob rem. 309558140 log. 17910
&c. &c.

&c.

&c.

25973 ift

rem.

134

6th.rem.

.

th

rem.

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