TO CALCULATE LOGARITHMS BY SERIES.

Les the garithm of 1 be required. Let 1+z = 1. Assume g. 1-z) = Az + Br2 + Cr3‚* &c., and, by a similar assumption, ing. 1+==A=+B=2+Cz3, &c. Bu: z=2z+n22 ̄z ̄ ̄z3, &c., and log. 1+z=n×log. 1+z =mås÷nBz3÷C3, &c.: therefore, by substituting the value of in the first expression of log. 1+z, and making it equal to the other, and then equating the coefficients, we get A=A, B——ja, C=+ja, D=-A; and therefore log.1+z = Ax (z — £x£ + £x3 — frt, &c.) Now log. a× (1 + z) = bg. ¢+kg. 1+1=log. a+A×(x —¿x2+}x3, &c.); and if az=y, or z=, log. (a+y)=log. a+Ax

(2−271+21, &c.) ; or if ar =—y, log. a—y=log. a —

2a

A×(++, &c.); and by subtraction, log.

In these series the quantity A is not determined, but it is a common multiplier of the series, and therefore is constant in the same system. If A =1, the system will be that of the natural or hyperbolic logarithms, which were the first invented by Lord Napier. Hence, in any other system, the logarithms may be got by multiplying the natural logarithms by the value of A in that other system. This value is called the module of the system.

TO FIND THE NATURAL LOGARITHM OF 10.

a must enter into every term of the series, that it may become when 0; and for the same reason r cannot be in the denominator. As n may denote any number, and the result is the same whatever its value is, it will be best to take n = 2; then log. 1+z = Ax+ B+ Cz3, &c. But = = x(2+x), therefore log. 1+= Ax(2+x)+ Bx2(2+x)2+Cx3(2+x)3, &c. = 2Ax+(A + 4B) x2 + (4B +8C)x3 +(B+12C+16D)x1, &c.; and it is also = 2Ax+2Bx2+2Cx3+2Dx1, &c. Here, by equating the coefficients of the same powers of x, we get AA, BA, &c. as in the text.

0

1

1

1+ + &c.) =•2231435513, &c. And making

3.81 5-8129

3

a 253

1024

1000

=

a+y a-y'

y = 12, = and

640095

9:

1000

3

=2X (1 +

253

9

+

3-64009 5-6400979 &c.)

5

=0237165266173, &c. Wherefore log. 10=10 log. +3

Because the common logarithm of 10 is 1, therefore 1 divided by 2.302585, &c. will give the module of the common logarithms 4342944819, &c.

=

Hence the natural logarithm multiplied by ·43429448, &c. will give the common logarithm; and the common logarithm multiplied by 2.30258, &c. will give the natural logarithm. To find the number belonging to a natural logarithm.

2.71828183, &c., the number of which the natural log. is 1. Required the common logarithms of the first 12 numbers.

USE OF LOGARITHMS IN EQUATIONS.

1. Let the equation be ab; then, in logarithms,

B

xAB, and x = where the capitals A, B, C, &c. repre

A'

sent the logarithms of the quantities a, b, c, &c.

2. Let =c; then mxA — (nx — 1) B — C, and x =

=d-P; then (n)B—mxC=(x —p)D, PD+nB±√(pD+nB|2 —4aB × (D+mC))

2

2D+2mC

G

PROBLEMS.

1. The duties on certain goods amounted to £2460, out of which a discount was allowed of 24 per cent. upon the sum actually paid for prompt payment. What did the discount amount to? Ans. £60.

2. A merchant discounted two bills at the bank, one of them for £550, payable in 7 months, and the other for £720, payable in 4 months; and he received for the whole £1200. At what rate per cent. per annum was the interest charged? Ans. £13.267 per cent. per annum.

3. The common difference of four numbers in arithmetical progression is 4, and their continual product is 21945. What are the numbers? Ans. 7, 11, 15, 19.

4. The sum of ten numbers in arithmetical progression is 120, and the sum of their cubes is 29160. What are the numbers? Ans. 3, 5, 7, 9, &c. 5. Given the sum of the numbers 0, 1, 2, 3, &c. = 1225; to find the sum of their squares. Ans. 40425.

6. Two persons set out at the same time from two places 462 miles distant, to meet one another. The first goes 1 mile the first day, 2 miles the second day, and so on. The second travels each day the cubes of the number of miles that the first travelled on that day. In what time will they meet? Ans. 6 days.

upon

7. A gentleman sold an estate for the value of the trees it above 7 feet circumference, at one pound for the first, two for the second, four for the third, and so on, doubling the price of each successive tree. The value of the estate came to £65535. How many trees of the above description were upon it? Ans. 16 trees.

8. A gentleman had seven children, whose ages differed successively by one year. In giving them new clothes, he determined to bestow as many yards of lace on the trimming of the youngest as he was years old, on the second as many as the sum of the ages of the two youngest, on the third as many as the sum of the ages of the three youngest, and so on; and he agreed with the tailor to pay for making each suit the product in pence of the child's age by the number of yards of lace on his suit. The bill amounted to £7, 10s. 6d. were the ages of the children? Ans. The youngest 5 years. 9. Required the number of combinations of m things in n things.

What

Ans. The number of combinations of two in n things is

n("71); of three, is n("71) ("~2); of four, is

3

=(" = 1) (" ~ 2 ) ( " — 3); of which the number of factors

s equal to the number of things in one combination. ThereFore the number of combinations of m in n things will be

(m-1)

m

10. A merchant discounted two bills; one had 6 months to run, and the other 8 months. The value of both came to £308, 6s. 8d., and the discount to £8, 6s. 8d. Had interest been charged upon the bills, it would have come to 4s. 83d. more than the discount. Required the value of the bills.

Ans. The bill due at 6 months was £205, and the other £103.

11. If £400 be the present value of an annuity to continue 23 years after the expiration of 8 years, what would be its value for 21 years after the expiration of 10 years, interest at 5 per cent.? Ans. £344-9597. 12. A gentleman had 10 different annuities of £100 each; their continuance differed by one year each, and the longest was for 60 years. He sold them all at 5 per cent. compound interest. What money did he receive for them?

Ans. £18653.26. 13. A bookseller purchases a work for £40, and pays for printing 1000 copies of it £15, for paper £20, and for incidents £10. He sells the edition in 10 years at 3s. each copy. How much does he gain per cent. per annum?

Ans. £11..19 per cent. per annum. 14. A person who owes his creditor £320 just now, and £96 more at the end of five years, wishes to pay the whole in one payment. What is the proper time for doing this, according to the true principle of equation of payments, viz. that the simple interest shall be equal to the discount?

Ans. At the end of one year.

15. A usurer lent £186 for a certain time, and gained £31; and by lending £360 at the same rate for another time, he gained £90. The sum of the times they were lent amounted to 20 months. How long time was each sum lent?

Ans. The first 8 months, the other 12 months.

PRACTICAL GEOMETRY.

DEFINITIONS.

1. GEOMETRY treats of magnitude or continued quantity, and of its relation to number.

2. A SOLID is that which has three dimensions, length, breadth, and thickness.

3. A SURFACE, or SUPERFICIES, is the boundary of a solid, and has only length and breadth.

4. A LINE is the boundary of a surface, and has only length.

5. A POINT is the extremity of a line. It has posi- .A tion, but not magnitude, as A.

6. A STRAIGHT LINE is uniform on all its sides. It can be exhibited by stretching a hair between two points, as AB.

7. A CURVE changes continually its di

rection, or it has unlike sides, a concave

and a convex, as CDE.

A

-B

D

E

8. An ANGLE is the measure of the relative position of two straight lines which meet, or it is their inclination to one another.

NOTE. An angle is denoted by three letters, of which the second is at the point where the lines meet, and the other two are upon the containing lines, one on each. Thus the uppermost angle is named ABC, the other CBD, B and the whole angle ABD.

9. A straight line is said to be PERPENDICULAR to another, when it does not incline towards one end more than towards the other. Thus AB is perpendicular to CD.

10. A RIGHT ANGLE is that made by a perpendicular, as CBG.

11. An OBTUSE ANGLE is greater than a right angle, as HKI.

C

-D

A

C

B

D

IG