c 2 
COMPOUND INTEREST. 
19 
.243 for £300, 
, the amount of 
shewing a dif 
fer the sum of 
in the time, 
years at 8 per 
under 2 per cent 
299 = £210 6 0 
30. To find (n) the number of years. 
(Art. 24.) log s = log p + mn X log fl + 
(Arith. and Alg. 109) by transposition, mn. log ( 1+ ^) = log s — logy; 
dividing each side by m. log ^1 + — ^ 
n __ log * — log p 
m X log ( 1 + — i 
\ m) 
Rule. Divide the difference of the logarithms of the principal and the 
amount by the logarithm of the sum to which £1 will amount at the 
first interval the interest is convertible, multiplied by the number of 
periods of conversion in the year. 
Example. In how many years will £210 6 0 amount to £690 at 
8 per cent compound interest payable quarterly ? 
p ~ 210.3 s ~ 690 i — .08 m — 4 
l° g (l + m) .00860017 
logs = 2.8388491 4_ 
logp~ 2.3228393 m. logfl + — ) = .03440068 
V m/ 
.03440068)0.5160098(15 years 
3440068 
1720030 
1720030 
210. 299 = P 
31, To find (m) the number of periods at which interest is convert 
ible in the year: 
s rr p 
Extracting the ?ith root. 
^ = (^—^) ’•, which equation there is no direct method of 
solving, but we can approximate sufficiently near by the following 
method : 
Expanding by the binomial theorem 
, . . . m — 1 .. m — 
14-i-4 i l 4- 
T T 2 m 2 m 
1 m — 2 
3 m 
m — 1 m — 2 m — 3 ^ ^ 
2 m ' 3 m 4rn \p