but if it be of a different name, take the difference of the indices, and prefix the sign of the greater.

EXAMPLES.

1. Required the quotient of .7591 divided by 32.147

Logarithm of .7591 is -1.88030

Do.

32.147 is

1.50714

Quotient .02361

Remain. -2.37316

In this example, the index of the divisor, with its sign changed, is -1, which added to -1, the index of the dividend, makes-2, for the index of the quotient.

2. Required the quotient of .63153 divided by .00917.

Logarithm of .63153 is -1.80039

Do.

.00917 is

-3.96237

Quotient 68.8683 Remain. 1.83802.

In this example there is 1 to carry from the decimal part of the logarithm, which subtracted from-3, the index of the divisor, leaves -2; this, with its sign changed, is+2; from which subtracting 1, the index of the dividend, the remainder is 1, and is affirmative, because the affirmative index is the greater.

3. Required the quotient of 13.921 divided by 7965.13

Logarithm of 13.921 is 1.14367

Do.

7965.13 is 3.90125

In this example there is 1 to carry from the decimal part of the logarithm, which added to 3, the index of the divisor, makes 4; this, with its sign changed, is, —4 ; from which subtracting 1, the index of the dividend, the remainder is -3.

4. Required the quotient of 79.35 divided by .05178. Ans. 1532.46.

5. Required the quotient of .5903 divided by .931. Ans. .63404.

PROBLEM V

To involve a number to any power, that is, to find the cube, &c. of a number, logarithmically.

RULE.

square,

Multiply the logarithm of the given number by the index of the power, viz. by 2 for the square, by 3 for the cube, &c. and the product will be the logarithm of the power.

Note. When the index of the logarithm is negative, if there be any to carry from the decimal part, instead of adding it to the product of the index and multiplier, subtract it, and the remainder will be the index of the logarithm of the power, and will always be negative.

EXAMPLES.

1. Required the square of 317.

Logarithm of 317

is 2.50106

2

4. Required the cube of 7.503. Ans. 422.37. 5. Required the 7th power of .32513. Ans..0003841

PROBLEM VI.

To extract any root of a number logarithmically.

RULE.

Divide the logarithm of the given number by the index of the root, that is, by 2 for the square root, by 3 for the cube root, &c. and the quotient will be the logarithm of the required root.

Note. When the index of the logarithm is negative, and does not exactly contain the divisor, increase the index by a number just sufficient to make it exactly divisible by it, and carry the units borrowed, as so many tens, to the left hand figure of the decimal part; then

Of the Arithmetical Complements of Logarithms.

When it is required to subtract several logarithms from others, it will be more convenient to convert the subtraction into an addition, by writing down, instead of the logarithms to be subtracted, what each of them

ing down what the first figure, on the right hand, wants of 10, and what every other figure wants of 9; this remainder is called the Arithmetical Complement. Thus, if the logarithm be 2.53061, its arithmetical complement will be 7.46939. If one or more figures to the right hand be ciphers, write ciphers in their place, and take the first significant figure from 10, and the remaining figures from 9. Thus, if the logarithm be 4.61300, its arithmetical complement will be 5.38700.

In any operation, where the arithmetical complements of logarithms are added to other logarithms, there must be as many tens subtracted from the sum, as there are arithmetical complements used.

As an example, let it be required to divide the product of 76.4 and 35.84, by the product of 473.9 and 4.76.