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CASE 2. When the dividend or divisor, or both of them, are decimal numbers.

RULE.

Subtract the decimal parts of the logarithms as before, and if 1 be borrowed in the left hand place of the decimal part, add it to the index of the divisor when that index is affirmative, but subtract it when negative.

Then conceive the sign of the index of the divisor changed from affirmative to negative, or from negative to affirmative; and if, when changed, it be of the same name with that of the dividend, add the indices together; 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

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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 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

Quotient .001748 Remain. -3.24242

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 square, cube, &c. of a number logarithmically.

RULE.

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

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.

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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 proceed with the division in the usual manner.

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5. Required the square root of 365. Ans. 19.105. 6. Required the 5th root of .9563. Ans. .9911. 7. Required the 4th root of .00079. Ans. .16765.

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 wants of 10.00000, which may readily be done, by writing 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 10's subtracted from the sum, as there are arithmetical complements used.

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