Elements of Algebra

180 ELEMENTS SECT. III. We have here $a = 3452$, and $n = 64$. Wherefore the logarithm of the amount sought is $64 \log. \frac{53}{50} + \log. 3452$, which is calculated thus: $log. 53 = 1.7242759$ subtracting $log. 50 = 1.6989700$ --- $log. \frac{53}{50} = 0.0253059$ multiplying by $64$ --- $64 \log. \frac{53}{50} = 1.6195776$ $log. 3452 = 3.5380708$ --- which gives $5.1576484$ And taking the number of this logarithm, we find the amount required equal to 143763l. 549. When the number of years is very great, as it is required to multiply this number by the logarithm of a fraction, a considerable error might arise from the logarithms in the Tables not being calculated beyond 7 figures of decimals; for which reason it will be necessary to employ logarithms carried to a greater number of figures, as in the following example. A principal of 1l. being placed at 5 per cent., compound interest, for 500 years, it is required to find to what sum this principal will amount, at the end of that period. We have here $a = 1$ and $n = 500$; consequently, the logarithm of the principal sought is equal to $500 \log. \frac{21}{20} + \log. 1$, which produces this calculation: $log. 21 = 1.322219294733919$ subtracting $log. 20 = 1.301029995663981$ --- $log. \frac{21}{20} = 0.021189299069938$ multiply by $500$ --- $500 \log. \frac{21}{20} = 10.594649534969000$, the logarithm of the amount required; which will be found equal to 39323200000l. 550. If we not only add the interest annually to the principal, but also increase it every year by a new sum $b$, the original principal, which we call $a$, would increase each year in the following manner: after 1 year, $\frac{21}{20}a + b$, after 2 years, $(\frac{21}{20})^2 a + \frac{21}{20}b + b$, after 3 years, $(\frac{21}{20})^3 a + (\frac{21}{20})^2 b + \frac{21}{20}b + b$,