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Here, the characteristic of log. 43218 being greater than that of log. 597.93, the remainder would be negative, to avoid which we add 10 to the log. of 597.93. The remainder is then the logarithm of a decimal fraction, and the position of the decimal point is determined by the rule in Art. 73.

77. As it is often more convenient to add than to subtract, the subtraction of logarithms is changed to addition by using the arithmetical complement.

The arithmetical complement of a logarithm is the remainder after subtracting it from 10.

The most convenient mode of finding it is to subtract the first significant figure on the right hand from 10, and all the others from 9. Thus, the ar. co. of 5.83116 is 4.16884

of 2.56467 is 7.43533

of 1.13450 is 8.86550 The arithmetical complement of the logarithm of a decimal fraction will be found by subtracting it from 20.

Thus, log. 0.08792 = 8.94409 and ar. co. = 11.05591
log. 0.00057 = 6.75587

= 13.24413

In performing division by logarithms, instead of subtracting the log. of the divisor we may add its arithmetical complement and reject 10 from the sum. For if b and c are any two numbers

b log.

-=log. b-log. c=log.b+10-log. C-10=

с

log. b + (ar. co.) log. C-10.

EXAMPLES.

To divide 1976.3 by 42.896.

log. 1976.3 = 3.29585 (ar. co.) log. 42.896 = 8.36758

log. 46.071 1.66343 The required quotient is therefore 46.071.

To divide 19.763 by 0.0042896.

log. 19.763 1.29585 (ar. co.) log. 0.0042896 12.36758

log 4607.1 3.66343 The required quotient is therefore 4607.1.

78. To involve a number to any power.

This is effected by Art. 62; that is, by multiplying the log. of the number by the exponent of the power.

EXAMPLES.

To find the fourth power of 5.7

log.
5.7 = 0.75587

4
log. 1055.6 = 3.02348
Therefore (5.7)4=1055.6.
To find the third power of 0.12
log. 0.12 = 9.07918

3 log. 0.001728 = 7.23754 Therefore (0.12)=0.001728.

Here log. 0.12 is 10 too great, and since we multiply by 3 the product must be 30 too great. But we reject only 20, in order to avoid a negative characteristic; it is therefore 10 loo great, and is the logarithm of a decimal fraction.

79. To extract any root of a number.

This is effected by Art. 63; that is, by dividing the log. of the number by the index of the root.

EXAMPLES.

To find the square root of 87825 log. 87825

ż | 4.94362

log. 296.35 = 2.47181 Therefore ✓87825 = 296.35.

If the logarithm has a negative characteristic, add as many times 10 as there are units in the index of the required root; the quotient will exceed its true value by 10.

To find the third root of 0.57894.

1130 + log. 0.57894 | 29.76263

log. 0.83345 = 9.92088 Therefore 70.57894 = 0.83345.

If we consider 9.76263 as the log. of 0.57894, we have to add 20 only, since it is already 10 too great.

80. To find the fourth term of a proportion.

The fourth term of a proportion is found by multiplying the second and third terms together, and dividing by the first. In logarithms this is effected by adding the logarithms of the second and third terms and subtracting the logarithm of the first term. Or instead of subtracting the log. of the first term, we may add its arithmetical complement. (Art. 77.)

EXAMPLES.

To find the fourth term of the proportion 33.16 : 398 :: 4.893 : X,

(ar. co.) log. 33.16 = 8.47939

log. 39 2.59988
log. 4.893 = 0.68958
log. 58.729

1.76885
Therefore x=58.729.
To find the fourth term of the proportion

0.0059768 : 478.3 :: 0.0076821 : X,

(ar. co.) log. 0.0059768

log. 478.3
log. 0.0076821

log. 614.77 Therefore x=614.77.

12.22353
2.67970
7.88548

2.78871

81. Complicated quantities of various kinds are readily computed by logarithms.

EXAMPLE.

To find the value of the fraction

7957x(410)*x(0.00379);

(4.653)*x(2.39) { log. 957=}| 2.98091

0.99364 410=2) 2.61278

5.22556 1 log. 0.00379=117.57864

8.78932 } log. 4.653 }| 0.66773=0.13355 (ar. co.) 9.86645 4 log. 2.39 - 41 0.37840=1.51360= (ar. co.) 8.48640

log. 2298.1

3.36137 Therefore the value of the fraction is 2298.1.

2 log.

4

1.

2.

82. The following examples will afford practice in all the usual operations by logarithms. (0.335)* X V18.6 x 5

=0.0069398.
897.65x(0.00017)
(25)+ x(26)*x(27)}

-2.9597.
(28)* x(29)+x(30)
(0.00015)= x(0.0000156)5 _5.5747
(587.69)8

(10)35
4. ✓(149.16) — (17.8)=148.09. (See Ex. 7, Art. 67.)

(567.8)3—(467.2) 5.

0.0011498. (15) X0.53

3.

6. ✓(49.755)2+(37.845)2 =62.513. (See Ex. 8, Art. 67.)

83. To solve an exponential equation by logarithms. In the exponential equation

ax=b,

log. 50
log. 8

if a=10, the value of x is found, by mere inspection, from a table of common logarithms, since in this case x is the common logarithm of b. But if a has any other value, as 8, then the value of x is the log. of b in the system whose base is 8, and may be found as follows. Take the equation

8x=50. Taking the logarithms of both members, we have, by. Art. 62,

x log. 8=log. 50,

1.69897 whence we have x=

1.8813.

0.90309 The division of 1.69897 by 0.90309 may also be performed by logarithms.

log. 1.69897=0.23019
log. 0.90309= 9.95573

log. 1.8813=0.27446 84. The complete solution of an exponential equation is therefore as follows: Taking the logarithms of the equation

ax=b, we have

a log. a = log.b,

[blocks in formation]

log. 6
log. a

and again taking the logarithms
log. x = log. =log. log. b-log. log. a.

log. a

log. 6

EXAMPLE

To find the value of x in the equation

1200 =100.
log. 100=2.00000, log. 2.00000=0.30103
log. 12=1.07918, log. 1.07918=0.03309

log. 1.8533

-0.26794 Therefore x=1.8533, or 121,8533. 100; that is, 1.8533 is the lo garithm of 100 in the system whose base is 12.

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