### Part I. Section III. Chapter 13. “Of the Calculation of Interest.”

540 We are accustomed to express the interest of any principal by per cents, signifying how much interest is annually paid for the sum of 100 pounds. And it is very usual to put out the principal sum at 5 per cent, that is, on such terms, that we receive 5 pounds interest for every 100 pounds principal. Nothing therefore is more easy than to calculate the interest for any sum; for we have only to say, according to the Rule of Three:

As 100 is to the principal proposed, so is the rate per cent to the interest required. Let the principal, for example, be 860l., its annual interest is found by this proportion;

100:5::860:43.

Therefore 43l. is the annual interest.

541 We shall not dwell any longer on examples of Simple Interest, but pass on immediately to the calculation of Compound Interest; where the chief subject of inquiry is, to what sum does a given principal amount, after a certain number of years, the interest being annually added to the principal. In order to resolve this question, we begin with the consideration, that 100l. placed out at 5 per cent, becomes, at the end of a year, a principal of 105l.: therefore let the principal be $$a$$; its amount, at the end of the year, will be found, by saying; As 100 is to $$a$$, so is 105 to the answer; which gives

$\dfrac{105a}{100} = \dfrac{21a}{20} = \frac{21}{20} \cdot a = a + \frac{1}{20}a.$

542 So that, when we add to the original principal its twentieth part, we obtain the amount of the principal at the end of the first year: and adding to this its twentieth part, we know the amount of the given principal at the end of two years, and so on. It is easy, therefore, to compute the successive and annual increases of the principal, and to continue this calculation to any length.

543 Suppose, for example, that a principal, which is at present 1000l., is put out at five per cent; that the interest is added every year to the principal; and that it were required to find its amount at any time. As this calculation must lead to fractions, we shall employ decimals, but without carrying them farther than the thousandth parts of a pound, since smaller parts do not at present enter into consideration.

Years Value (l.)
after 1 year 1050
52.5
after 2 years 1102.5
55.125
after 3 years 1157.625
57.881
after 4 years 1215.506
60.775

which sums are formed by always adding ¹⁄₂₀ of the preceding principal.

544 We may continue the same method, for any number of years; but when this number is very great, the calculation becomes long and tedious; but it may always be abridged, in the following manner:

Let the present principal be $$a$$, and since a principal of 20l. amounts to 21l. at the end of a year, the principal $$a$$ will amount to $$\frac{21}{20}\cdot a$$ at the end of a year: and the same principal will amount, the following year, to $$\frac{21^2}{20^2}\cdot a=\left(\frac{21}{20}\right)^2 \cdot a$$. Also, this principal of two years will amount to $$\left(\frac{21}{20}\right)^3 \cdot a$$, the year after: which will therefore be the principal of three years; and still increasing in the same manner, the given principal will amount to $$\left(\frac{21}{20}\right)^4 \cdot a$$ at the end of four years; to $$\left(\frac{21}{20}\right)^5 \cdot a$$, at the end of five years; and after a century, it will amount to $$\left(\frac{21}{20}\right)^{100} \cdot a$$; so that, in general, $$\left(\frac{21}{20}\right)^n \cdot a$$ 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 ²¹⁄₂₀, which is used in this calculation, depends on the interest having been reckoned at 5 per cent., and on ²¹⁄₂₀ being equal to ¹⁰⁵⁄₁₀₀. But if the interest were estimated at 6 per cent, the principal $$a$$ would amount to $$\frac{106}{100}\cdot a$$, at the end of a year; to $$\left(\frac{106}{100}\right)^2\cdot a$$, at the end of two years; and to $$\left(\frac{106}{100}\right)^n \cdot a$$, at the end of $$n$$ years.

If the interest is only at 4 per cent, the principal $$a$$ will amount only to $$\left(\frac{104}{100}\right)^n \cdot a$$, after $$n$$ years.

546 When the principal $$a$$, as well as the number of years, is given, it is easy to resolve these formulae by logarithms. For if the question be according to our first supposition, we shall take the logarithm of $$\left(\frac{21}{20}\right)^n \cdot a$$, which is

$=\log \left(\frac{21}{20}\right)^n + \log a;$

because the given formula is the product of $$\left(\frac{21}{20}\right)^n$$ and $$a$$. Also, as $$\left(\frac{21}{20}\right)^n$$ is a power, we shall have $$\log \left(\frac{21}{20}\right)^n=n \log \frac{21}{20}$$: so that the logarithm of the amount required is $$n \log \frac{21}{20} + \log a$$; and farther, the logarithm of the fraction $$\frac{21}{20}$$ equals $$\log 21 - \log 20$$.

547 Let now the principal be lOOOl. 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:

Subtracting

$\begin{array}{rr} \log 21&=1.3222193\\ \log 20&=1.3010300\\ \hline \log \frac{21}{20}&=0.0211893. \end{array}$

Multiplying by 100

$100\log \frac{21}{20} = 0.021189300.$

Adding $$\log 1000$$,

$\begin{array}{rr} 100\log \frac{21}{20}&=0.021189300\\ \log 1000&=3.0000000\\ \hline 100\log \frac{21}{20}+\log 1000&=5.1189300 \end{array}$

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?

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:

Subtracting

$\begin{array}{rr} \log 53&=1.7242759\\ \log 50&=1.6989700\\ \hline \log \frac{53}{50}&=0.0253059. \end{array}$

Multiplying by 64

$64\log \frac{53}{50} = 1.6195776.$

Adding $$\log 3452$$,

$\begin{array}{rr} 64\log \frac{53}{50}&=1.6195776\\ \log 3452&=3.5380708\\ \hline 64\log \frac{53}{50}+\log 3452&=5.1576484 \end{array}$

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:

Subtracting

$\begin{array}{rr} \log 21&=1.322219294733919\\ \log 20&=1.301029995663981\\ \hline \log \frac{21}{20}&= 0.021189299069938 \end{array}$

Multiplying 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 $$\left(\frac{21}{20}\right)^2a$$ $$+\frac{21}{20}b$$ $$+b$$ after 3 years $$\left(\frac{21}{20}\right)^3a$$ $$+\left(\frac{21}{20}\right)^2b$$ $$+\frac{21}{20}b$$ $$+b$$ after 4 years $$\left(\frac{21}{20}\right)^4a$$ $$+\left(\frac{21}{20}\right)^3b$$ $$+\left(\frac{21}{20}\right)^2b$$ $$+\frac{21}{20}b$$ $$+b$$ after $$n$$ years $$\left(\frac{21}{20}\right)^na$$ $$+\left(\frac{21}{20}\right)^{n-1}b$$ $$+\left(\frac{21}{20}\right)^{n-2}b$$ $$+\cdots$$ $$+\frac{21}{20}b$$ $$+b$$

This amount evidently consists of two parts, of which the first is $$\left(\frac{21}{20}\right)^na$$; and the other, taken inversely, forms the series

$b+\frac{21}{20}b+\left(\frac{21}{20}\right)^2b+\left(\frac{21}{20}\right)^3b+\cdots+ \left(\frac{21}{20}\right)^{n-1}b;$

which series is evidently a geometrical progression, the ratio of which is equal to ²¹⁄₂₀, and we shall therefore find its sum, by first multiplying the last term $$\left(\frac{21}{20}\right)^{n-1}b$$ by the exponent ²¹⁄₂₀; which gives $$\left(\frac{21}{20}\right)^nb$$. Then, subtracting the first term $$b$$, there remains $$\left(\frac{21}{20}\right)^nb-b$$; and, lastly, dividing by the exponent minus 1, that is to say by ¹⁄₂₀, we shall find the sum required to be $$20\left(\frac{21}{20}\right)^nb-20b$$; therefore the amount sought is,

$\left(\frac{21}{20}\right)^na + 20\left(\frac{21}{20}\right)^nb-20b = \left(\frac{21}{20}\right)^n \cdot(a+20b)-20b.$

551 The resolution of this formula requires us to calculate, separately, its first term $$\left(\frac{21}{20}\right)^n \cdot (a+20b)$$, which is $$n\log \frac{21}{20}+\log(a+20b)$$; for the number which answers to this logarithm in the Tables, will be the first term; and if from this we subtract $$20b$$, we shall have the amount sought.

552 A person has a principal of 1000l. placed out at five per cent, compound interest, to which he adds annually 100l. beside the interest: what will be the amount of this principal at the end of twenty-five years?

We have here $$a=1000$$; $$b=100$$; $$n=25$$; the operation is therefore as follows:

$\log \frac{21}{20} = 0.021189299$

Multiplying by 25, we have

$25 \log \frac{21}{20} = 0.5297324750$ $\log(a+20b)=3.4771213135$

And the sum is

$25 \log \frac{21}{20}+\log(a+20b) = 4.0068537885.$

So that the first part, or the number which answers to this logarithm, is 10159.1, and if we subtract $$20b = 2000$$, we find that the principal in question, after twenty-five years, will amount to 8159.1l.

553 Since then this principal of 1000l. is always increasing, and after twenty-five years amounts to 8159⅒l. we may require, in how many years it will amount to 1000000l.

Let $$n$$ be the number of years required: and, since $$a=1000$$, $$b=100$$, the principal will be, at the end of $$n$$ years, $$\left(\frac{21}{20}\right)^n \cdot (3000) - 2000$$, which sum must make 1000000; from it therefore results this equation;

$3000\cdot \left(\frac{21}{20} \right)^n -2000 = 1000000;$

And adding 2000 to both sides, we have

$3000\cdot \left(\frac{21}{20}\right)^n = 1002000.$

Then dividing both sides by 3000, we have $$\left(\frac{21}{20}\right)^n = 334$$.

Taking the logarithms, $$n\log \frac{21}{20} = \log 334$$; and dividing by $$\log \frac{21}{20}$$, we obtain $$n=\dfrac{\log 334}{\log \frac{21}{20}}$$. Now, $$\log 334=2.5237465$$, and $$\log \frac{21}{20} = 0.0211893$$; therefore $$n=\dfrac{2.5237465}{0.0211893}$$; and, lastly, if we multiply the two terms of this fraction by 10000000, we shall have $$n=\dfrac{25237465}{211893}$$ = 119 years, 1 month, 7 days; and this is the time, in which the principal of 1000l. will be increased to 1000000l.

554 But if we supposed that a person, instead of annually increasing his principal by a certain fixed sum, diminished it, by spending a certain sum every year, we should have the following gradations, as the values of that principal $$a$$, year after year, supposing it put out at 5 per cent, compound interest, and representing the sum which is annually taken from it by $$b$$:

 after 1 year $$\frac{21}{20}a$$ $$-b$$ after 2 years $$\left(\frac{21}{20}\right)^2a$$ $$-\frac{21}{20}b$$ $$-b$$ after 3 years $$\left(\frac{21}{20}\right)^3a$$ $$-\left(\frac{21}{20}\right)^2b$$ $$-\frac{21}{20}b$$ $$-b$$ after $$n$$ years $$\left(\frac{21}{20}\right)^na$$ $$-\left(\frac{21}{20}\right)^{n-1}b$$ $$-\left(\frac{21}{20}\right)^{n-2}b$$ $$-\cdots$$ $$-\frac{21}{20}b$$ $$-b$$

555 This principal consists of two parts, one of which is $$\left(\frac{21}{20}\right)^n \cdot a$$, and the other, which must be subtracted from it, taking the terms inversely, forms the following geometrical progression:

$b+\frac{21}{20}b+\left(\frac{21}{20}\right)^2b+\left(\frac{21}{20}\right)^3b+\cdots+ \left(\frac{21}{20}\right)^{n-1}b;$

Now we have already found (Art. 550) that the sum of this progression is $$20\left(\frac{21}{20}\right)^nb-20b$$; if, therefore, we subtract this quantity from $$\left(\frac{21}{20}\right)^n \cdot a$$, we shall have for the principal required, after $$n$$ years,

$\left(\frac{21}{20}\right)^n \cdot (a-2b)+20b.$

556 We might have deduced this formula immediately from that of Art. 550. For, in the same manner as we annually added the sum $$b$$, in the former supposition; so, in the present, we subtract the same sum $$b$$ every year. We have therefore only to put in the former formula, $$-b$$ everywhere, instead of $$+b$$. But it must here be particularly remarked, that if $$20b$$ is greater than $$a$$, the first part becomes negative, and, consequently, the principal will continually diminish. This will be easily perceived; for if we annually take away from the principal more than is added to it by the interest, it is evident that this principal must continually become 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 $$-20000\left(\frac{21}{20}\right)^n + 120000$$, or $$120000-20000\left(\frac{21}{20}\right)^n$$, where the factor, -20000, is the result of $$a-20b$$; or 100000 - 120000.

So that the principal will become nothing, when $$20000\left(\frac{21}{20}\right)^n$$ amounts to 120000; or when $$20000\left(\frac{21}{20}\right)^n = 120000$$. Now, dividing both sides by 20000, we have $$\left(\frac{21}{20}\right)^n=6$$; and taking the logarithm, we have $$n \log \frac{21}{20} = \log 6$$; then dividing by $$\log \frac{21}{20}$$, we have $$n=\dfrac{\log 6}{\log \frac{21}{20}}$$, or $$n=\dfrac{0.7781513}{0.0211893}$$; 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 show how, from the same principles, we may calculate interest for times shorter than whole years. For this purpose, we make use of the formula $$\left(\frac{21}{20}\right)^n\cdot a$$ 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 $$\left(\frac{21}{20}\right)^{\frac{8}{365}} \cdot 100000$$; the logarithm of which quantity is

$\log \left(\frac{21}{20}\right)^{\frac{8}{365}}+\log 10000=\frac{8}{365} \log \frac{21}{20} + \log 100000.$

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

$\frac{8}{365}\log \frac{21}{20}+\log 100000=5.0004644$

The natural number of this logarithm is found to be 100107. So that, subtracting the principal, 100000 from this amount, the interest, for eight days, is 107l.

560 To this subject belong also the calculation of the present value of a sum of money, which is payable only after a term of years. For as 20l., in ready money, amounts to 21l. in a year; so, reciprocally, a sum of 21l., which cannot be received till the end of one year, is really worth only 20l. If, therefore, we express, by $$a$$, a sum whose payment is due at the end of a year, the present value of this sum is $$\frac{20}{21}a$$; and therefore to find the present worth of a principal $$a$$, payable a year hence, we must multiply it by ²⁰⁄₂₁; to find its value two years before the time of payment, we multiply it by $$\left(\frac{20}{21}\right)^2a$$; and in general, its value, $$n$$ years before the time of payment, will be expressed by $$\left(\frac{20}{21}\right)^na$$.

561 Suppose, for example, a man has to receive for five successive years, an annual rent of 100l. and that he wishes to give it up for ready money, the interest being at 5 per cent; it is required to find how much he is to receive. Here the calculations may be made in the following manner:

For 100l. due

 after 1 year, he receives 95.239 after 2 years, he receives 90.704 after 3 years, he receives 86.385 after 4 years, he receives 82.272 after 5 years, he receives 78.355 Sum of the 5 terms = 432.955

So that the possessor of the rent can claim, in ready money, only 432.955l.

562 If such a rent were to last a greater number of years, the calculation, in the manner we have performed it, would become very tedious; but in that case it may be facilitated as follows:

Let the annual rent be $$a$$, commencing at present and lasting $$n$$ years, it will be actually worth

$a+\left(\frac{20}{21}\right)a+\left(\frac{20}{21}\right)^2a+\left(\frac{20}{21}\right)^3a+\left(\frac{20}{21}\right)^4a+\cdots+\left(\frac{20}{21}\right)^na.$

Which is a geometrical progression, and the whole is reduced to finding its sum. We therefore multiply the last term by the exponent, the product of which is $$\left(\frac{20}{21}\right)^{n+1}a$$; then, subtracting the first term, there remains $$\left(\frac{20}{21}\right)^{n+1}a-a$$; and, lastly, dividing by the exponent minus 1, that is, by -¹⁄₂₁, or, which amounts to the same, multiplying by -21, we shall have the sum required,

$-21\cdot \left(\frac{20}{21}\right)^{n+1}\cdot a + 21a,$

or,

$21a-21\cdot \left(\frac{20}{21}\right)^{n+1}\cdot a;$

and the value of the second term, which it is required to subtract, is easily calculated by logarithms.

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.