and last figure of the required number-the proportional part for 7 being, as we have seen, (not 120, but) 130. EXAMPLE I. Find the product of 456700, 23.895, and *000834756. Adding together 5.6596310, 1.3783070, and 4'9215595,-the logarithms of 456700, 23-895, and 000834756, respectively,we find the logarithm of the required product to be 3'9594975; so that the product itself is 9109:56: 3'9594975=log 9109.56; 9109*56=456700 X 23·895 X 000834756. Log 31664 1864=4'5005684 Log 785'4 =2.8950909 EXAMPLE II.-Divide 31664 1864 by 785'4. Subtracting 2 8950909, the logarithm of 7854,from 4'5005684, the logarithm of 31664 1864, we find the logarithm of the required quotient to be 16054775; so that the quotient itself is 40316-31664 1864-7854. 40.316. 1-6054775 16054775=log 40316; EXAMPLE III.—Find the fourth term (a) of the proportion 25348: 19327.56:: 385.694: a Adding together 4.2861770 and 2.5862429, the logarithms of the means, we find the logarithm of the product of the means to be 6.8724199, from which we subtract 3 4039437, the logarithm of the given extreme. The remainder, 3.4684762, is the logarithm of the fourth term of the proportion; so that the fourth term (a) itself is 2940-8728. Log 19327′56 =4·2861770 6.8724199 Log 25348=3*4039437 3'4684762 34684762-log 2940 8728; 2940.8728=a. NOTE.-In finding the difference between two logarithms, we sometimes begin by adding to both a number which makes the subtrahend 10. The number so added is known as the "arithmetical complement" of the subtrahend, and is, of course, the difference between the subtrahend and 10. Thus, the arithmetical complement of 3.4039437 being (10-3*4039437=) 6.5960563,* the subtraction of 10 from 13.4684762—that is, of 34039437+65960563 from 6·8724199+6·5960563—leaves the same remainder as the subtraction of 3.4039437 from 6.8724199. So that the logarithm of the fourth term of the last proportion can be obtained as follows, 10 being subtracted mentally from 13-the integral part of the sum of the three addends: Log 85764= T·9333050 EXAMPLE IV. Find the 17th power of 85764. The logarithm of 85764 being 19333050, the logarithm of 85764" is 19333050X17=2·8661850; so that the required power is 073483 (nearly). 17 Log 85764"-2.8661850 2.8661850=log 073483; 07348385764". In multiplying 19333050 by 17, we obtain (9333050X17=) 15.8661850, which is additive, from the mantissa; and (I×17=) 17, which is subtractive, from the characteristic. We therefore write the product (17+15.8661850) under the form 2.8661850; 17+15 being equivalent to 2. Because, diminishing a number by 2 is obviously the same as performing the twofold operation of (a) adding 15 to the number, and (b) subtracting 17 from the result. EXAMPLE V.-Extract the 7th root of 316.4758. The logarithm of 316-4758 being 2.5003405, the logarithm of 316-4758 is of 2.5003405i.e., 03571915; so that the required root is 2.2761. Log 316'4758=2.5003405; 2.5003405÷7=0'3571915; 0*3571915=log 2.2761; 2.27613164758 7 EXAMPLE VI.-Extract the 5th root of '000000813972. The logarithm of '000000813972 being 7.9106095, the logarithm of 000000813972 is of 79106095-i.e., 2.7821219; so that the required root is 060551 : * This is most easily obtained when we subtract the last figure (7) of 3'4039437 from 10, and each of the others from 9-the "carrying" of I being dispensed with. The manner in which 7.9106095 is divided by 5 requires a word of explanation. One-fifth of 7·9106095 being 1.5821219, one-fifth of 79106095 would, at first sight, appear to be 15821219; but if we multiply 1.5821219 by 5, we shall obtain (5+2·9106095=) 3·9106095, instead of 7.9106095. The division of 7 by 5 would give I for quotient, and leave 2-equivalent to 20 tenths-for remainder. If 9 tenths were added to this remainder, the result would be (not 29, but) II tenths, the division of which by 5 would give 2 for quotient, and leave 'I for remainder. So that, if the division were continued, the mantissa would be marked minus, as well as the characteristic ; whereas A MANTISSA IS NEVER MARKED MINUS* (unless when written, as a subtrahend, after another logarithm). In every such case as the one under consideration, we begin by so altering the form of the logarithm as to have, for characteristic, a multiple of the divisor; and we naturally select the lowest multiple which answers our purpose. The first multiple of 5, after 7, being 10, which exceeds 7 by 3, we convert 7.9106095 into 10+3'9106095 by adding 3 to the characteristic, and +3 to the mantissa; 3+3 being=0.† We then find, with the greatest facility, that of 7·9106095 is 2.7821219. 4.7689317 Again: in dividing 47689317 by 3, we first add 2 to the characteristic, and +2 to the mantissa; the lowest multiple of 3, after 4, being 6, which exceeds 4 by 2; and 2+2 being =0. We then find that of 4.7689317 is 2.9229772. 3)6+2.7689317 2.9229772 * This explains why the sign minus is always placed above (instead of before) the characteristic of the logarithm of a decimal. If placed before the characteristic, the sign would be understood to indicate that the whole of the logarithm was subtractive. It is obvious that if 9106095 were added to a number, and 7 subtracted from the sum, the result would be exactly the same as if 39106095 were added to the number, and 10 subtracted from the sum. 57 7, we find that the fraction lies between and, and that, consequently, it is more accurate to substitute for than to reject altogether; so that, as a fourth and still closer approximation, we may set down I 2+ or (the terms of the admits of no alteration, its numerator being 1, the work of 66 decomposition "—as it is sometimes called-terminates here; and when is taken into account, we necessarily obtain, instead of another approximation, the original fraction 485 itself: I 122 Of such approximations as the foregoing,—each closer than the preceding one, the first is too large; the second, too small; the third, too large; the fourth, too small; and so on, alternately. Thus, in writing for 485, which is really equiva lent to I II22 we take a fraction whose denominator is too small; so that the first approximation is too large. Again, ¦ 2+ 29 31 485 ; so that the second approximation is too small. More II22 I over, being a larger fraction than 3+ is a larger num 529 I I fraction than—5+ is a larger number than 5+47 5+1 |