Unaccompanied by explanation, the results of the preceding divisions and subtractions may be written in this way: 6)409(68.16; 11.361; 18935185 [3 divisions.] 18935185)6(31687042; 16734477 [2 divisions.] 1-6734477)1 8935185(11315074 [1 division.] 1*1315074)1*6734477 (14789543; 1*3070655;} [4 divisions.] 10208985)11315074(1*1083447; 1.0856561;) [6 divisions.] 10634319; 10416627; 10203391; If 292-87=205; 205-87=118; 118–87=31 [3 subtractions.] 87-31=56; 56-31=25_[2 subtractions.] 31-25-6 [1 subtraction.] 25-6-19; 19-6-13; 13-6-7; 7-6=1 [4 subtractions.] 6—1=5; 5—1=4; 4—1=3; 3—1=2; 2—1=1; 1—1=0 [6 subtractions.] It will be seen that, in working this series of subtractions, we proceed exactly as if, without the aid of division, we desired to find the greatest common measure of 292 and 87-subtracting 87 as often as possible from 292 being the same as dividing 292 by 87; subtracting 31 as often as possible from 87, the same as dividing 87 by 31; and so on: By means of the "quotients" 3, 2, 1, 4, 6, we can at once determine 292 the index of the power which 409 is of 6; and 87 we find these quotients by treating 409 and 6 in the manner just explained, and noting the number of divisions (3, 2, 1, 4, and 6, respectively), performed with each divisor: 6 being first divided as often as possible into 409; 18935185, the last of the resulting quotients, being next divided as often as possible into 6; 16734477, the last of the second set of quotients, being then divided as often as possible into 1.8935185; and so on-the process being continued until I is obtained for quotient: Knowing that the index of the power which 409 is of 6 must be greater than unity, we set down 3, the first "quotient," as the integral part of the mixed number to which the index is reducible: we then find, in the usual way, that the fractional part is 3, and that, consequently, the index is 337, or 297. So that, if the base were 6, the logarithm of 409 would be 3.3563218. Next, let it be required to find the "Common" logarithm of 5: in other words, to express 5 as a power of 10. Proceeding as in the last case, we obtain the following results : 5)10(2. [1 division.] 2)5(25; 125. [2 divisions.] 125)2(1.6; 128; 1'024. [3 divisions.] 1024) 125(12207031; 1*1920929; 1·1641532; 11368684; 1'1102230; 10842021; 10587911; 10339757; 10097419. [9 divisions.] 1*0097419) 1924 (10141205; 10043363. [2 divisions] 10043363)10097419(10053822; 1'0010414. [2 divisions.] 1*0010414) 10043363 (1·0032915; 1'0022477; 10012050; 10001634. [4 divisions.] 10001634) 10010414 (10008778; 10007143; 1*0005508; 10003873; 1'0002238; 10000604. [6 divisions.] 1*0000604) 1*0001634(1·0001030; 10000426. [2 divisions.] 1*0000426) 1*0000604(1*0000178. [1 division.] 1*0000178) 1*0000426 (10000248; 10000070. [2 divisions.] 1*0000070) 1*0000178(10000108; 10000038. [2 divisions.] 1*0000038) 1*0000070(1*0000032. [1 division.] 1*0000032) 1*0000038(10000006. [1 division.] The difference between this last quotient and I being almost inappreciable, the divisions terminate here; and the index of the power which 5 is of 10-in other words, the logarithm of 5-is found, as follows, to be 453043 or '6989700: 648158' COMPOUND INTEREST. 296. Compound interest-which is commonly, but erroneously, defined to be "interest on interest -may be regarded as consisting of two parts, namely: (a) interest on principal, and (b) interest on interest. If £1,000 were lent for 3 years, at 5 per cent., and the interest paid in yearly instalments, the lender would receive, as interest on principal,—or as simple interest,—at the end of the In addition to this, the lender would receive £7 12s. 6d.—interest on interest-by investing, as principal (at 5 per cent.), the first 50 for the last two years, the second £50 for the third year, and a further sum of £2 10s. for the third year; i.e., the £210s. falling due on the first £50 at the end of the second year. So that the compound interest on £1,000 for 3 years, at 5 per cent., would be (£150+£7 12s. 6d.=) £157 12s. 6d.: £ s. = 2 10 = 2 10 3rd = 2 10 Ist £50 for 2nd year 3rd Interest on 026 This result can be obtained in a different way. Remembering that the "amount" (i.e., the principal+the interest) for any particular year is the "principal" available for the following year, we have £ 1,000 0 s. d. principal 1,157 12 6 amount 1,000 O o=original principal 157 12 6 =compound interest. It is hardly necessary to observe that the calculation of compound interest in either of the preceding ways would be extremely tedious and troublesome particularly if the time for which the principal was lent extended over a great many years. But compound interest can be calculated more easily. If I were invested; as principal, at (say) 5 per cent., on the understanding that the "interest on principal" should fall due in annual instalments, the first year's interest would be £05, and the first year's amount (£1+05=) £105. This amount would become the second year's principal; so that two principals being proportional to their respective amounts for equal periods of time, when the rate per cent. is the same in both cases the second year's amount would be the last term of the proportion (Ist (Ist (2nd (2nd year's prin.) year's amt.) year's prin.) year's amt.) LI : £105 £105 : £1052 As 1052, the second year's amount, would become the third year's principal, the third year's amount would be the last term of the proportion (Ist (Ist (3rd (3rd year's prin.) year's amt.) year's prin.) year's amt.) As 1053, the third year's amount, would become the fourth year's principal, the fourth year's amount would be the last term of the proportion (Ist (Ist (4th (4th year's prin.) year's amt.) year's prin.) year's amt.) £1.051 It could be shown, in the same manner, that the fifth year's |