CHAPTER XXXIX.
INTEREST AND ANNUITIES.
450. Questions involving Simple Interest are easily solved by the rules of Arithmetic; but in Compound Interest the calculations are often very laborious. We shall now show how these arithmetical calculations may be simplified by the aid of logarithms. Instead of taking as the rate of interest the interest on $100 for one year, it will be found more convenient to take the interest on $1 for one year. If this be denoted by $r, and the amount of $1 for 1 year by $R, we have R=1+r.
451. To find the interest and amount of a given sum in a given time at compound interest.
Let P denote the principal, R the amount of $1 in one year, n the number of years, I the interest, and M the amount.
The amount of P at the end of the first year is PR; and, since this is the principal for the second year, the amount at the end of the second year is PR x R or PR^2 Similarly the amount at the end of the third year is PR^3 and so on; hence the amount in n years is PR^n; that is,
M=PR^n and therefore I=P(R^n-1).
Ex. Find the amount of $100 in a hundred years, allowing compound interest at the rate of 5 per cent, payable quarterly ; having given log 2 = .8010300, log 3 = .4771213, log 14.3906 = 1.15808.
The amount of $1 in a quarter of a year is $(1+ 1 4 5 100) or $81 8).
