Elements of Algebra

CHAP. XIII. OF ALGEBRA. 179 principal will amount to $(\frac{21}{20})^4 . a$ at the end of four years; to $(\frac{21}{20})^5 . a$, at the end of five years; and after a century, it will amount to $(\frac{21}{20})^{100} . a$; so that, in general, $(\frac{21}{20})^n . a$ will be the amount of this principal, after $n$ years; and this formula will serve to determine the amount of the principal, after any number of years. 545. The fraction $\frac{21}{20}$, which is used in this calculation, depends on the interest having been reckoned at 5 per cent., and on $\frac{21}{20}$ being equal to $\frac{105}{100}$. But if the interest were estimated at 6 per cent, the principal $a$ would amount to $\frac{106}{100} . a$, at the end of a year; to $(\frac{106}{100})^2 . a$, at the end of two years; and to $(\frac{106}{100})^n . a$, at the end of $n$ years. If the interest is only at 4 per cent. the principal $a$ will amount only to $(\frac{104}{100})^n . a$, after $n$ years. 546. When the principal $a$, as well as the number of years, $n$, is given, it is easy to resolve these formulæ by logarithms. For if the question be according to our first supposition, we shall take the logarithm of $(\frac{21}{20})^n . a$, which is $= log. (\frac{21}{20})^n + log. a$; because the given formula is the product of $(\frac{21}{20})^n$ and $a$. Also, as $(\frac{21}{20})^n$ is a power, we shall have $log. (\frac{21}{20})^n = n log. \frac{21}{20}$: so that the logarithm of the amount required is $n log. \frac{21}{20} + log. a$; and farther, the logarithm of the fraction $\frac{21}{20} = log. 21 - log. 20$. 547. Let now the principal be $1000l$. and let it be required to find how much this principal will amount to at the end of 100 years, reckoning the interest at 5 per cent. Here we have $n = 100$; and, consequently, the logarithm of the amount required will be $100 log. \frac{21}{20} + log. 1000$, which is calculated thus: log. 21 = 1.3222193 subtracting log. 20 = 1.3010300 log. $\frac{21}{20}$ = 0.0211893 multiplying by 100 $100 log. \frac{21}{20} = 2.1189300$ adding log. 1000 = 3.0000000 gives 5.1189300 which is the logarithm of the principal required. We perceive, from the characteristic of this logarithm, that the principal required will be a number consisting of six figures, and it is found to be $131501l$. 548. Again, suppose a principal of $3452l$. were put out at 6 per cent, what would it amount to at the end of 64 years? N