Elements of Algebra

CHAP. XIII. OF ALGEBRA. 183 come less, and at last it will be absolutely reduced to nothing; as will appear from the following example: 557. A person puts out a principal of 100000l. at 5 per cent interest; but he spends annually 6000l.; which is more than the interest of his principal, the latter being only 5000l.; consequently, the principal will continually diminish; and it is required to determine, in what time it will be all spent. Let us suppose the number of years to be $n$, and since $a = 100000$, and $b=6000$, we know that after $n$ years the amount of the principal will be $-20000 (\frac{21}{20})^n + 120000$, or $120000 - 20000(\frac{21}{20})^n$, where the factor, $-20000$, is the result of $a-20b$; or $100000 - 120000$. So that the principal will become nothing, when $20000(\frac{21}{20})^n$ amounts to $120000$; or when $20000(\frac{21}{20})^n = 120000$. Now, dividing both sides by $20000$, we have $(\frac{21}{20})^n = 6$; and taking the logarithm, we have $n \log. (\frac{21}{20}) = \log. 6$; then dividing by $\log. \frac{21}{20}$, we have $n = \frac{\log. 6}{\log. \frac{21}{20}}$, or $n = \frac{0.7781513}{0.0211893}$; and, consequently, $n = 36$ years, 8 months, 22 days; at the end of which time, no part of the principal will remain. 558. It will here be proper also to shew how, from the same principles, we may calculate interest for times shorter than whole years. For this purpose, we make use of the formula $(\frac{21}{20})^n . a$ already found, which expresses the amount of a principal, at 5 per cent, compound interest, at the end of $n$ years; for if the time be less than a year, the exponent $n$ becomes a fraction, and the calculation is performed by logarithms as before. If, for example, the amount of a principal at the end of one day were required, we should make $n = \frac{1}{365}$; if after two days, $n = \frac{2}{365}$, and so on. 559. Suppose the amount of 100000l. for 8 days were required, the interest being at 5 per cent. Here $a = 100000$, and $n = \frac{8}{365}$, consequently, the amount sought is $(\frac{21}{20})^{\frac{8}{365}} \times 100000$; the logarithm of which quantity is $\log. (\frac{21}{20})^{\frac{8}{365}} + \log. 100000 = \frac{8}{365} \log. \frac{21}{20} + \log. 100000$. Now, $\log. \frac{21}{20} = 0.0211893$, which, multiplied by $\frac{8}{365}$, gives $0.0004644$, to which adding $\log. 100000 = 5.0000000$ the sum is $5.0004644$.