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4. To divide 0.7438 by 12.9476.
Here, the 1 taken from I, gives 2 for a result, as set
13. The Arithmetical complement of a logarithm is the number which remains after subtracting the logarithm from 10.
14. We will now show that, the difference between two logarithms is truly found, by adding to the first logarithm the arithmetical complement of the logarithm to be subtracted, and then diminishing the sum by 10.
10 – 9.274687 = 0.725313.
0.725313 is the arithmetical complement
Let a= the first logarithm,
b= the logarithm to be subtracted,
c10-6 the arithmetical complement of b.
Now the difference between the two logarithms will be expressed by a b.
But, from the equation c=10-b, we have
hence, if we place for b its value, we shall have
which agrees with the enunciation.
When we wish the arithmetical complement of a loga rithm, we may write it directly from the table, by subtracting the left hand figure from 9, then proceeding to the right, subtract each figure from 9 till we reach the last figure, which must be taken from 10: this will be the same as taking the logarithm from 10.
1. From 3.274107 take 2.104729.
By common method.
Hence, to perform division by means of the arithmetical complement, we have the following
its ar. comp.
To the logarithm of the dividend add the arithmetical com plement of the logarithm of the divisor: the sum, after sub tracting 10, will be the logarithm of the quotient.
Divide 0.875 by 25.
By arith. comp.
Sum 1.169378 after sub
In this example, the sum of the characteristics is 8, from which, taking 10, the remainder is 2.
3. Divide 37.149 by 523.76.
log 523.76 ar. comp. 7.280867
FINDING THE POWERS AND ROOTS OF NUMBERS BY LOGARITHMS.
15. We have (Art. 3),
10m = M.
Raising both members of this equation to the nth power, we have,
in which mXn is the logarithm of M" (Art. 1): hence, The logarithm of any power of a given number is equal to the logarithm of the number multiplied by the exponent of the power.
16. Taking the same equation,
10" — M,
and extracting the nth root of both members, we have
10" = M
in which is the logarithm of M: that is,
The logarithm of the root of a given number is equal to the logarithm of the number divided by the index of the root.
1. What is the 5th power of 9?
Log 90.954243; 0.954243 x 5 = 4.771215; whole number answering to 4.771215 is 59049.
2. What is the 7th power of 8?
3. What is the cube root of 4096?
Log 4096 3.612360; 3.612360÷ 3 = 1.204120; number answering to 1.204120 is 16.
4. What is the 4th root of .00000081?
Log .00000081 = 7.908485;
8 +1.908485 ÷ 4 = 2.477121,
the number answering to which is .03, which is the rooi When the characteristic of the logarithm is negative, and not divisible by the index of the root, add to it such a negative number as will make the sum exactly divisible by the index, and then prefix the same number to the first decimal figure of the logarithm. 5. What is the 6th root of .0432? Ans. .592353 +. 6. What is the 7th root of .0004967? Ans. .3372969,
17. Before explaining the method of constructing geometrical problems, we shall describe some of the simpler intruments and their uses.
18. The dividers is the most simple and useful of the instruments used for drawing. It consists of two legs ba, bc, which may be easily turned around a joint at b.
One of the principal uses of this instrument is to lay off on a line, a distance equal to a given line.
For example, to lay off on CD a distance equal to AB.
ger on the joint of the dividers, and A-
the thumb and other fingers, the с other leg of the dividers, until its foot reaches the point B. Then raise the dividers, place one foot at C, and mark with the other the distance CE: this will evidently be equal to AB.
RULER AND TRIANGLE.
19. A Ruler of convenient size, is about twenty inches in length, two inches wide, and a fifth of an inch in thick
It should be made of a hard material, perfectly straight and smooth.
The hypothenuse of the right-angled triangle, which is used in connection with it, should be about ten inches in length, and it is most convenient to have one of the sides considerably longer than the other. We can solve, with the ruler and triangle, the two following problems.
I. To draw through a given point a line which shall be parallel to a given line.
20. Let be the given point, and AB the given line. Place the hypothenuse of the triangle against the edge of the ruler, and then place the ruler and triangle. on the paper, so that one of the sides of the triangle shall coincide exactly with AB: the triangle being below the line.
Then placing the thumb and fingers of the left hand firmly on the ruler, slide the triangle with the other hand along the ruler until the side which coincided with AB reaches the point C. Leaving the thumb of the left hand on the ruler, extend the fingers upon the triangle and hold it firmly, and with the right hand, mark with a pen or pencil, a line through C: this line will be parallel to AB.
II. To draw through a given point a line which shall be perpendicular to a given line.
21. Let AB be the given line, and D the given point. Place the hypothenuse of the tri
angle against the edge of the ruler, as before. Then place the ruler and triangle so that one of the sides of the triangle shall coincide exactly with the line AB. Then slide the triangle along the ruler until the other side reaches the point D: draw through D a right line, and it will be perpendicular to AB.