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When subtracting logarithms, their algebraic difference is to be found. The operation may sometimes be confusing, because the mantissa is always positive, and the characteristic may be either positive or negative. When the logarithm to be subtracted is greater than the logarithm from which it is to be taken, or when negative characteristics appear, subtract the mantissa first, and then the characteristic, by changing its sign and adding.

EXAMPLE.-Divide 274.2 by 6,784.2.

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First subtracting the mantissa .83150 gives .60657 for the mantissa of the quotient. In subtracting, 1 had to be taken from the characteristic of the minuend, leaving a characteristic of 1. Subtract the characteristic 3 from this, by changing its sign and adding 1 - 3 = 2, the characteristic of the quotient. Number corresponding to 2.60657 = .040418. Hence, 274.2 ÷ 6,784.2 .040418.

=

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1, number cor

responding to 1.39343 = 24.742. Hence, .067842.002742 =

24.742.

The only case that is likely to cause trouble in subtracting is that in which the logarithm of the minuend has a nega tive characteristic, or none at all, and a mantissa less than the mantissa of the subtrahend. For example, let it be required to subtract the logarithm 3.74036 from the logarithm

3.55145. The logarithm 3.55145 is equivalent to -3+.55145. Now, if we add both +1 and -1 to this logarithm, it will not change its value. Hence, 3.55145—3—1+1+.55145 — - 4 +1.55145. Therefore, 3.55145-3.74036 =

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Had the characteristic of the above logarithm been 0 instead of 3, the process would have been exactly the same. Thus, .55145 1+1.55145; hence,

=

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Hence,

Number corresponding to 4.60657 = .00040417.

.0274267.842 .00040417.

EXAMPLE.-What is the reciprocal of 3.1416?

SOLUTION.-Reciprocal of 3.1416

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1

1

and log

3.1416'

3.1416

= 0.49715. Since 0 = -1+1,

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INVOLUTION BY LOGARITHMS.

If X represents a number whose logarithm is x, we have, from the definition of a logarithm,

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Raising both numbers to some power, as the nth, the equation becomes

10** = X".

But X is the required power of X, and xn is its logarithm, from which it follows that the logarithm of a number

multiplied by the exponent of the power to which it is raised is equal to the logarithm of the power. Hence, to raise a number to any power by the use of logarithms:

Rule.-Multiply the logarithm of the number by the exponent that denotes the power to which the number is to be raised, and the result will be the logarithm of the required power.

EXAMPLE.-What is (a) the square of 7.92? (b) the cube

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of 94.7? (c) the 1.6 power of
SOLUTION.-(a) Log 7.92
Hence, .89873 × 2 = 1.79746
ing to 1.79746 62.727. Hence, 7.922
(b) Log 94.7
Number corresponding to 5.92905
94.73 = 849,280, nearly.

(c) Log 5121.6

=

=

=

1.97635; 1.97635 X 3 = 5.92905 = log 94.73. 849,280, nearly. Hence,

=

1.6 X log 512 = 1.6 X 2.70927 = 4.334832, or 4.33483 (when using five-place logarithms) Hence, 5121.6

=

21,619 nearly.

=

log 21,619.

If the number is wholly decimal, so that the characteristic is negative, multiply the two parts of the logarithm separately by the exponent of the number. If, after multiplying the mantissa, the product has a characteristic, add it, algebraically, to the negative characteristic multiplied by the exponent, and the result will be the negative characteristic of the required power.

EXAMPLE.-Raise .0751 to the fourth power.
SOLUTION.-Log.07514

=

4 X log .0751

=

4X2.87564.

tiplying the parts separately, 4×2

=

Mul

8 and 4 X .87564

3.50256. Adding the 3 and 8, 3 + (— 8) = -5; therefore, log .07514 5.50256. Number corresponding to this

=

.00003181. Hence, .07514 = .00003181.

A decimal may be raised to a power whose exponent contains a decimal as follows:

EXAMPLE.-Raise .8 to the 1.21 power.

SOLUTION.-Log .81.21 = 1.21 × 1.90309. There are several ways of performing the multiplication.

First Method.-Adding the characteristic and mantissa. algebraically, the result is -.09691. Multiplying this by 1.21 gives-.1172611, or -.11726, when using five-place logarithms. To obtain a positive mantissa, add +1 and -1; whence, log .81.21-1 +1.11726 1.88274.

Second Method.-Multiplying the characteristic and mantissa separately gives −1.21 + 1.09274. Adding characteristic and mantissa algebraically, gives —.11726; then, adding +1 and -1, log .81.21 1.88274.

=

Third Method.-Multiplying the characteristic and mantissa separately gives -1.21 +1.09274. Adding the decimal part of the characteristic to the mantissa gives −1+(-.21 +1.09274) = 1.88274 = log.81.21. The number corresponding to the logarithm 1.88274 = .76338.

Any one of the above three methods may be used, but we recommend the first or the third. The third is the most elegant and saves figures, but requires the exercise of more caution than the first method does. Below will be found the entire work of multiplication for both .81.21 and .8.21.

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In the second case, the negative decimal obtained by multiplying -1 and .21 was greater than the positive decimal obtained by multiplying .90309 and .21; hence, +1 and -1 were added, as shown.

EVOLUTION BY LOGARITHMS.

If X represents a number whose logarithm is x, we have, from the definition of a logarithm,

10*
= X.

Extracting some root of both members, as the nth, the

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x

But X is the required root of X, and is its logarithm,

n

from which it follows that the logarithm of a number divided

root to be extracted is equal to the Hence, to extract any root of a number

by the index of the
logarithm of the root.
by means of logarithms:

Rule.-Divide the logarithm of the number by the index of the root; the result will be the logarithm of the root.

EXAMPLE.-Extract (a) the square root of 77,851; (b) the cube root of 698,970; (c) the 2.4 root of 8,964,300.

SOLUTION.-(a) Log 77,851

= 4.89127; the index of the root

is 2; hence, log 77,851 = 4.89127 ÷ 2 corresponding to this

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279.02. Hence, V 77,851

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698,970
88.747, nearly.

(c) Log28,964,300

or,28,964,300

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= 6.95251 2.4 = 2.89688 log 788.64; 788.64, nearly.

If it is required to extract a root of a number wholly decimal, and the negative characteristic will not exactly contain the index of the root, without a remainder, proceed as follows:

Separate the two parts of the logarithm; add as many units (or parts of a unit) to the negative characteristic as will make it exactly contain the index of the root. Add the same number to the mantissa, and divide both parts by the index. The result will be the characteristic and mantissa of the root. EXAMPLE.-Extract the cube root of .0003181. SOLUTION.-Log.0003181

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=

log.0003181
3

4.50256

6) + (2 + .50256 = 2.50256).

2)+(2.50256 ÷ 3 = .83419);

log .0003181 2.83419 log .068263.

.0003181.068263.

EXAMPLE.-Find the value of .0003181.

SOLUTION.-Log1.0003181

3

1.41

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If -.23 be added to the characteristic, it will contain 1.41 exactly 3 times. Hence,

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