28 
ON THE VALUE OF ANNUITIES. 
p == 260.433 
By Table 3, 
* = .08 
(1.02) 24 = 1.608437 
334.062 
n rr 6 
m rr 4 
3216874 
965062 
6434 
482 
48 
418.8900 = £418 17 10 
By logarithms, 
log 1.02 rr 0.00860017 
24 
3440068 
1720034 
log 1.02 24 = 0.20640408 
log p = 2.4156960 
2.6221001' 418.890 
42. To find (//) the number of years. 
Substituting in the formula of Art. 37, the logarithm of the amount 
of ¿£l when interest is payable m times a year for the logarithm of the 
amount when interest is payable yearly, it becomes, 
Rule. Divide the difference of the logarithms of the present value 
and the sum due, by the logarithm of the amount of £l at the end of 
the first interval, multiplied by the number of intervals. 
Example. ¿£260 8 8 is paid down in lieu of ¿£350, 6 per cent 
compound interest payable half-yearly being allowed as discount. How 
long was the sum paid before due ? 
p rr 260.433 s ~ 350 
i — 06 
log s 2.5440680 log 
log p r= 2.4156958 
02567444)0.1283722(5 years 
log 1.03 = 0.01283722 
2 
.02567444 
1283722 
43. To find (i) the rate of interest. 
Substituting in the formula of Art. 38, the number of intervals for 
the number of years, we have the interest for one interval;