will be obtained; which being denoted by S, we shall log. b = log. a + 1 log. c + 1 S. have log.c 2 lcg. b. log. a= log. a- S. As an example hereof, let it be proposed to find the hyperbolic logarithm of 3. Then the hyperbolic logarithm of 2 being already found =,693147180, &c. that of 4, which is the double thereof, will also be known. Therefore, taking a = 2, b = 3, and c =4, we shall, in this case, have x 1 (2ac + 1) = 17, x2 = Therefore 4S (~+~~~+~-~, &c.) =,058891517, &c. and hyp. log. 2 + hyp. log. 4. 2.5 (x 3 5 consequently hyp. log. 3. = 1,098612288, &c. 2 2. Let the hyberbolic logarithm of 10 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 1 2dc + being= 1 161' we shall have S (x + = + **, &c.) = =,005211180, &c. +,000000079, &c. 3 5 &c. =,006211259, &c. And therefore log. 10 (2 log. 9 2,302585092, &c. - log. 8 S) = 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 a unit is assumed for the logarithm of 10, the logarithm of any number will be found, by multiplying 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, as 2,302585092, &c. the hyperbolic log. of 10 above found is to (H) the hyperbolic logarithm of any other number, so is 1, H X the common logarithm of 10, to H (2,302585092, &c. ,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 already shown), and which, by making R,868588963, Rx Rx 3 &c. will stand more commodiously thus, Rx + + 5 For an example hereof, let the common logarithm of 7 be required: in which case the logarithms of 8 and 9 being known, from those of 2 and 3, we shall have log. 7 2 log. 8-log. 9 Rx3 Rx5 S (by the Theor.), S being = Rx + + &c. (= the common log. of 64), and x (= 3 5 1 127: whence 64 63 64 +63 Rx3 Rx5 Consequently S (Rx+ + &c.) =,006839424, 3 5 &c. and 2 log. 8 log. 9 ·S =,845098040, &c. = the common logarithm of 7 required. But the same conclusion may be brought out by fewer terms of the series, if the logarithms of the three first primes 2, 3, and 5 be supposed known; because those of 48 and 50 (which are composed of them) will likewise be known; from whence the log. 48.+log. 50 + S logarithm of 7 (= log. 49 = 4 + S) will come out =,845098040, &c. (as before), which value will be true to 11 places of figures, by taking the first term of the series, only. Again, let the common logarithm of the next prime number, which is 11, be required. Here a may be taken = 10, b = 11, and c = 12; but fewer terms of the series will suffice, if other three numbers, composed of 11 and the inferior primes be taken, whereof the common difference is a unit. Thus, because 982 X 7 X 7, 99 3 x 3 x 11 (9 × 11), and 100 = 2 × 2 × 5 x 5 (or 10 x 10), 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, notwithstanding all these artifices and compendiums, a method, similar to that in page 21, for finding the logarithms of large numbers, one from another, by addition and subtraction only, still seems wanting in the calculation of tables; I shall, 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 part of the difference of the logarithms of A and B; then the sum will be the excess of the logarithm of A +1 above that of A. 4. From this excess let the quotient, found by Rule 2, be continually subtracted, that is, first from the excess itself, then from the remainder, then from the next remainder, &c. 5. To the logarithm of A add the said excess, and to the sum add the first of the remainders; to the last sum add the next remainder, &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. Rule 2.), which multiplied by 49,5, and the product added log. Blog. A to 100 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: |