It thus appears that 1.1315074 can be divided into 1.6734477 as often-and only as often-as 6 can be subtracted from 25.

Substituting 1-1315074 for d, we have

Now, as often as I can be subtracted from 6, so often, exactly, ought 10208985 to be contained as factor in 1131507+; and this we find to be the case:

Unaccompanied by explanation, the results of the preceding divisions and subtractions may be written in this way:

* This last quotient would be exactly if 409 were exactly the 292nd power of the 87th root of 6; but the index 292 is only an 87 approximation.

6)409(68.16; 11.361; 18935185 [3 divisions] 1.8935185)6(3*1687042; 1·6734477 [2 divisions] 1-6734477)1-8935185(1°1315074 [1 division] 1.1315074)16734477(14789543; 13070655; [4 divi11551542; 10208985 sions] 1*0208985)11315074(11083447; 10856561; [6 divi10634319; 10416627; 10203391; 15 sions] 292-87=205; 205-87=118; 118- [3 subtrac

87=315

tions]

87-31-56; 56-31-25 [2 subtractions.]

31-25-6 [1 subtraction.]

6=1 S

[4 subtractions]

[6 subtrac

25-6-19; 19-6=13; 13-6=7; 7

6-1=5;5-1=4; 4-1=3; 3-1=2; \ 2-1=1; 1-1=0

tions]

In working the 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; &c. :

By means of the "quotients" 3, 2, 1, 4, 6, we can at

once determine

292, the index of the
87

of 6; and we obtain 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; 1·6734477, the last of the second set of quotients, being then divided as often as possible into 18935185; &c.-the process being continued until 1 is obtained for quotient :

66

2

14

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, and that, consequently, the index is 384, or 292. So that, if the base were 6, the logarithm of 409 would be 3·3563218.

:

Next, let it be required to find the 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(2·5; 125 [2 divisions]

1*25)2(1·6; 1·28; 1024 [3 divisions]

1024)1*25(1*2207031; 1*1920929; 1*1641532 11368684; 11102230; 10842021; 10587911;

10339757; 1'0097419.

[9 divi

sions]

1*0097419) 1·024(1°0141205; 1'0043363 [2 divisions] 1*0043363) 1*0097419(1*0053822; 1'0010414 [2 divisions] 10010414)1*0043363(10032915; 10022477; [4 divi1*0012050; 10001634 sions] 10001634)10010414(1*0008778; 10007143; [6 divi1.0005508; 1.0003873; 10002238; 10000604 sions]

1*0000604) 1*0001634(10001030; 10000426 [2 divs.] 10000426) 10000604(10000178 [1 division] 1.0000178) 1*0000426(1·0000248; 1·0000070 [2 divs.] 1*0000070) 1*0000178(10000108; 10000038 [2 divs.] 1*0000038) 10000070(10000032 [1 division] 1*0000032) 10000038(1.0000006 [1 division]

The difference between this last quotient and 1 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 thus found to be 453043, or '6989700: 648158'

I

2

3

9 2

2

6

2

I

2

2

I

I

COMPOUND INTEREST.

297. 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 (at compound interest) for 3 years, at 5 per cent. per annum, 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. per annum), 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 £2 108. 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. per annum, would be (150+£7 128. 6d.=) £157 12s. 6d. :

This result can be obtained in a different way. Remembering that the "amount" (i.e., the principal+the

=157 12

6