Here in subtracting I suppose 10 to be added to the first characteristic, and say 8 from 11, &c.

In taking the arithmetical complement, the logarithm of the number may be subtracted immediately from 10. The logarithm of 2 being .301030, its arithmetical complement is T.698970. Adding 10 it becomes 9.698970. It would be the same if subtracted immediately from 10 thus 10--.301030 9.698970.

8. It is required to find the value of x in the following expression.

9.

Find the value of x in the following equations.

Observe that the 2d power of 38 is found by multiplying the logarithm of 38 by 2, the 3d power by multiplying it by 3, &c. which will give the logarithm of the result. Hence we have the following equation; the logarithm of 38 being 1.579784 and that of 583 being 2.765669.

The value of x is found by dividing one logarithm by the other in the same manner as other numbers. It might be done by logarithms if the tables were sufficiently extensive to take out the numbers. By a table with six places an answer correct to four decimal places may be obtained.

In taking out the logarithms the right hand figure may be omitted without affecting the result in the first four decimals.

log. 2.76567

log. 1.57978

0.441800

0.198596

log. x = 1.75067

0.243204

13. What is the value of x in the equation 1537

= 52?

This gives first 1537 = 52o.

This may now be solved like the last.

LII. Questions relating to Compound Interest:

It is required to find what any given principal p will amount to in a number n of years, at a given rate per cent. r, at compound interest.

Suppose first, that the principal is $1, or £1, or one unit of money of any kind.

The interest of 1 for one year is

1 X r

100

or simply r, if ris considered a decimal. The amount of 1 for one year then, will be 1+r. The amount of p dollars will be p (1+r).

For the second year, p (1+r) will be the principal, and the amount of 1 being (1+r), the amount of p (1+r) will be p (1 + r) (1 + r) or p (1+r)2.

For the third year p (1+r)2 being the principal, the amount will be p(1+r)2 (1 + r) or p (1 + r)3.

For n years then, the amount will be p (1+r)".

1

Putting A for the amount, we have

A = p(1+r)".

This equation contains four quantities, A, p, r, and n, any three of which being given, the other may be found.

Logarithms will save much labour in calculations of this

kind.

Examples.

1. What will $753.37 amount to in 53 years, at 6 per cent. compound interest?

2. What principal put at interest will amount to $5000 in 13 years at 5 per cent. compound interest?

3. At what rate per cent. must $378.57 be put at compound interest, that it may amount to $500 in 5 years?

Solving the equation A = p (1+r)" making r the unknown quantity, it becomes

4. In what time will $284.37 amount to 750 at 7 per cent.? Making n the unknown quantity, the equation A = p(1+r)"

5. What will be the compound interest of $947 for 4 years and 3 months at 5 per cent.?

6. What will $157.63 amount to in 17 years at 4 per cent.?

7. A note was given the 15th of March 1804, for $58.46, at the rate of 6 per cent. compound interest; and it was paid the 19th of Oct. 1823. To how much had it amounted?

8. A note was given the 13th of Nov. 1807, for $456.33, and was paid the 23d of Sept. 1819. The sum paid was $894.40. What per cent. was allowed at compound interest?

9. In what time will the principal p be doubled, or become 2 p, at 6 per cent. compound interest? In what time will it be tripled?

Note. In order to solve the above question, put 2 p in the place of A for the first, 3 p for the second, and find the value of n.

The principles of compound interest will apply to the following questions concerning the increase of population.

10. The number of the inhabitants of the United States in A. D. 1790 was 3,929,000, and in 1800, 5,306,000. What rate per cent. for the whole time ws the increase? What per cent. per year?