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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; 1.8935185, 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 1.8935185; and so on—the process being continued until 1 is obtained for quotient : 3 2

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4 514

31 87

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 337, or 237. 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(2-5; 1'25. [2 divisions.]
1-25)2(1.6; 1'28; 1'024. [3 divisions. ]

1°024) 1*25(1·2207031 ; 1'1920929; 1'1641532 ; 1'1368684 ; I•1102230; 1.0842021 ; 1.0587911; 1'0339757 ; I*0097419. [9 divisions.]

1'0097419) 1924(10141205; 1'0043363. [2 divisions] 1'0043363)1.0097419(1.0053822 ; 1'0010414. [2 divisions.]

1*0010414)10043363( 140032915; 1'0022477 ; 1'001 2050 ; I'0001634. [4 divisions.]

I'0001634) 1'0010414 (1'0008778; 1'0007143 ; 1'0005508 ; 1'0003873 ; 1'0002238 ; 1'0000604. [6 divisions.]

1'0000604) 1'0001634(1*0001030 ; 1'0000426. [2 divisions. ] 1'0000426)1*0000604(1'0000178. [1 division.] 1'0000178) 1'0000426(1'0000248; 1.0000070. [2 divisions.] 1'0000070)10000178(1'0000108 ; 1'0000038. [2 divisions.] 1'0000038)1'00000701'0000032. [1 division.] I '0000032)1*0000038(1'0000006. [ı 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:

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Putting for the required index, we have

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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

£50
2nd
3rd

Ist year,

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£150 In addition to this, the lender would receive £7 128. 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 £2 nos. 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+ £7 128. 60.=) £157 128. 6d.:

£ s. d.
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Interest on

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6 principal = 1500 Compound interest

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6 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

£ 3. d. 1,000

principal
50

interest
= Ist year's
1,050

amount

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amount
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( principal
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interest
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1,157 12
6

amount
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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(£i+£.055) £1:05. 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

(1st (Ist (2nd (2nd year's prin.) year's amt.) year's prin.) year's amt.)

£1 £105 :: £105 : £1.052 As £105, 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.)

£1 : £105 :: £105 : £1053 As £1:05, 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

(1st (4th (4th year's prin.) year's amt.) year's prin.) year's amt.)

£1 : £105 :: £1053 : £1:05 It could be shown, in the same manner, that the fifth year's

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amount would be £105); the sixth year's, £ 105®; the seventh year's, £1:05’; &c.:

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If the interest on principal fell due in half-yearly instalments, the first half-year's interest would be £'025, and the first halfyear's amount (£1+£:025=)£1.025. This amount would become the second half-year's principal. So that, from a series of proportions like the preceding, the second half-year's amount would be found to be £ 1.0252 ; the third half-year's, £1°0253; the fourth half-year's, £1.0254 ; &c. :-

£1
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amount
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£
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principal

amount
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amount
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£1025}=3rd half-year's

=4th half year's { principal

If the interest on principal fell due in quarterly instalments, the first quarter's interest would be £-0125, and the first quarter's amount (£1+£0125=) £10125. This amount would become the second quarter's principal. So that the second quarter's amount would be £1.0125%; the third quarter's, £1.01233; the fourth quarter's, 210125*; &c. :

£1•0125, 1=2nd quarter's

£1
£1'0125
=ist quarter's

S principal

amount

principal
£101252

amount
£10125
£1.01253
=3rd quarter's principal

amount
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principal Z10125"}=4th quarter's

amount &c.

&c.

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