We shall reduce the first 24 of the terms in this series to decimal expressions, carrying each to 25 figures, then add them together, and multiply the sum by 2, which, as the above equation denotes, will give the logarithm of 2. Here also, as in finding the base of Napier's system, the student would be aided by Hutton's table of reciprocals.

1÷3 .333 333 333 333 333 333 333 333 3 1÷(3.33)=.012 345 679 012 345 679 012 345 7 1÷(5.35)=.000 823 045 267 489 711 934 156 4 1 (7.37) = .000 065 321 052 975 373 963 028 3 1(9.39) .000 005 645 029 269 476 762 237 0 1÷(11.311)=.000 000 513 184 479 043 342 021 6 1÷(13.313)=.000 000 048 248 113 414 331 301 2 1÷(15.315)=.000 000 004 646 114 625 083 754 9 1÷(17.317)=.000 000 000 455 501 433 831 740 7 1÷(19.319)=.000 000 000 045 283 768 275 670 2 1÷(21.321)=,000 000 000 004 552 336 493 321 3 1÷(23.323)=.000 000 000 000 461 831 238 452 9 1÷(25,32 5)=.000 000 000 000 047 209 415 486 3 1÷(27.327)=.000 000 000 000 004 856 935 749 6 1÷(29.329)=.000 000 000 000 000 502 441 629 3 1÷(31.331)=.000 000 000 000 000 052 225 115 6 1÷(33.33 3)=.000 000 000 000 000 005 451 106 3 1÷(35.335)=.000 000 000 000 000 000 571 068 3 1÷(37.337)=.000 000 000 000 000 000 060 022 2 1÷(39.339)=.000 000 000 000 000 000 006 327 1 1÷ ÷(41.341)=.000 000 000 000 000 000 000 668 7 1 (43.343)=.000 000 000 000 000 000 000 070 8 1÷(45.345)=.000 0000 000 000 000 000 000 007 5 1÷(47.347)=.000 000 000 000 000 000 000 000 8 .346 573 590 279 972 654 708 616 0 =

half the logarithm of 2, and .346 573 590 279 972 654 708616 0×2 = .693 147 180 559 945 309 417 232 1, the loga. rithm of 2, correct to 25 places of decimals

In order to find the logarithm of 3 we shall proceed with

same manner as with the above series for finding the loga. rithm of 2, first reducing the several terms of the series to decimal expressions, and then adding these expressions together.

1÷2 .500 000 000 000 000 000 000 000 0 1÷(3.23)=.041 666 666 666 666 666 666 666 6 1÷(5.25)=.005 250 000 000 000 000 000 000 0 1÷(7.27)=.001 116 071 429 571 428 571 428 5 1(9.29)=.000 217 013 888 888 888 888 888 8 1÷(11.211)=.000 044 389 204 545 454 545 454 5 1÷(13.213)=.000 009 390 024 038 461 538 461 5

÷(15.215)=.000 002 034 505 208 333 333 333 3 1÷(17.217)=.000 000 448 787 913 602 941 176 5 1 (19.219)=.000 000 100 386 770 148 026 315 8 1÷(21.221)=.000 000 022 706 531 343 005 952 3 1 (23.223)=.000 000 005 183 012 589 164 402 2 1÷(25.225)=.000 000 001 192 092 885 507 812 5 1÷(27.227)=.000 000 000 275 947 429 515 697 5 1÷(29.229)=.000 000 000 064 229 143 076 929 5 1÷(31.231)=.000 000 000 015 021 331 848 636 7 1..(33.223)=.000 090 000 003 224 706 722 028 3 1÷(35.235)=.000 000 000 000 831 538 913 049 5 1÷(37.237)=.000 000 000 000 196 647 503 086 1 1÷(39.239)=.000 000 000 000 046 640 753 929 4

1÷(41.241)=.000 000 009 000 011 091 398 802 1 1÷(43.243)=.000 000 000 000 002 643 879 947 0 1÷(45.245)=.000 000 000 000 000 631 593 542 9 1÷(47.247)=.000 000 000 000 000 151 179 305 5 1 (49.249) .000 000 000 000 000 036 252 180 4 1÷(51.251)=.000 000 000 000 000 008 707 631 6 1÷(53.25 3)=.000 000 000 000 000 002 094 760 4 1÷(55,255)=.000 000 000 000 000 000 504 646 9

1 (57.257)=.000 000 000 000 000 000 121 638 5 1÷(59.259)=.000 000 000 000 000 000 029 378 8 1÷(61.261)=.000 000 000 000 000 000 007 103 9 1÷(63.263)=.000 000 000 000 000 000 001 719 6 1÷(65.265)=.000 000 000 000 000 000 000 416 7. 1÷(67.267)=.000 000 000 000 000 000 000 101 0 1÷(69,269)=.000 000 000 000 000 000 000 024 5 1÷(71.271)=.000 000 000 000 000 000 000 005 9 1÷(73.273)=.000 000 000 000 000 000 000 001 5 1÷(75.275)=.000 000 000 000 000 000 000 000 3 .549 306 144 334 054 845 697 622 6 = half the logarithm of 3, and .549 306 144 334 054845 697 622 6×2=1.098 612 288 668 109 691 395 245 2, the loga. rithm of 3, correct to 25 places of decimals.

The logarithm of 2 was found to be .693 147 180 559 945309 417 232 1, which multiplied by 2, or added to itself, becomes 1.386 294 361 119 890 618 834 464 2, the logarithm of 4.

Note. The logarithms of 2, 3 and 4, as we have computed them are certainly the result of no inconsiderable labor; and the extent to which they are carried, can be required of a student only in extreme cases. The several answers given by Day are carried to only 6 places of decimals, which will generally be sufficient unless great accuracy is sought. In the remaining numbers whose logarithms are required, the decimal places will be carried only so far as is necessary to obtain the given answers.

1

1

Log. 5=log. 3+2) +. ·+. + , &c.)

1
3.43 5.45 7.47

,י

=

=.510826 and added to 1.098612, the logarithm of 3,= 1.609438, the logarithm of 5.

The logarithm of 6=log. 2+log. 3=.693147+1.098612 =1.791759.

Log. 7=log. 5+2 + + +

1

1
1
3.63 5.65 7.67

-.1666666

,י

&c.

1 1

-- =

6 6

.1682360, which multiplied by 2=

.3364720, and added to 1.609438, the logarithm of 51.945910, the logarithm of 7.

The logarithm of 8=the logarithm of 2 multiplied by 3, or added to itself 3 times .693147+.693147+.693147=

2.079441.

The logarithm of 9=the logarithm of 3 added to itself= 1.098612+1.098612-2.197224.

The logarithm of 10 the logarithm of 2 added to the logarithm of 5.693147+1.609438=2.302585.

COMPUTATION OF COMMON OR BRIGGS'S LOGARITHMS.

According to Day, in order to find the logarithm of any number in the common system, it is necessary first to find the value of the modulus, which (Art. 67,) is equal to 1 divided by Napier's logarithm of 10; that is 12.302585 = .43429448. This number substituted for M, or twice the number, viz. .86658896, substituted for 2M in the series in Art. 68, will enable us to calculate the logarithm of any number in the common or Briggs' system.

Although the above method has its value, still we believe

that the logarithms of numbers according to Briggs' system
can be computed by a more simple process, when it is our ob-
ject to derive them directly from Napier's, or from those com-
puted according to any other system. By Art. 67, the ratio of
the logarithms of two numbers to each other, is the same in one
system as in another; so that if we have the logarithms of
numbers in one system, and the logarithm of any number
given in Briggs' system, we can find the logarithm of any
other number in the common system. For example, log. 10
in Napier's system: log. 2 in the same system::log. 10 in
Briggs's system: log. 2 in the same system. It may be ob.
jected that we have not the log. of any number in Briggs'
system given. But we simply take the log. of his base or
radix, which we are to consider as given at the very com-
mencement of our calculations. To proceed with our meth-
od of calculation, we will first find the log. of 2: then,

Napier's log. 10 Napier's log. 2: : Briggs's log. 10:
Briggs's log 2:

or, .693 147 180 559 945 309 417 232 1 : 2.302 585 092-
994 045 684 017 991 4::1; x. Multiplying extremes and
means, .693 147 180 559 945 309 417 232 1x =2.302 585.
092 994 045 684 017 991 4, and x =301 029 995 663 981-
195 213 738, the logarithm of 2, according to Briggs' sys-
tem, to 24 places of decimals. To go through with the pro-
cess of division which is necessary in order to obtain the re-
sult, would, indeed, be tedious, but the operation becomes
comparatively easy when we wish to obtain the logarithm
to only 6 or 7 places of decimals, so that the truth of the
previous remark in reference to the comparative simplicity
of this method and that of Day, remains unimpaired. In or
der, then, to obtain the common logarithm of any number
when we have the logarithms of the numbers in any other
system given, we have the following rule:

In the given system, divide the logarithm of any number by the logarithm of 10, and the quotient will be the logarithm of that number according to the required system, whose logarithm is the dividend. According to this rule the log. of 2, 3, 4, 5, 6, 7, 8, 9, 10, are as follows:

Log. 2=2.302585÷.693147=.301030
Log. 3=2.302585 1.098612.477121
Log. 4 2.302585÷1.386294.602060
Log.5=2.302585÷1.609438.698970