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.

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

66

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.

log.

For if b and c are any two numbers

b с

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

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

EXAMPLES.

To divide 1976.3 by 42.896.

log. 1976.33.29585 (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.

=

To find the fourth power of 5.7 5.7

log.

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.

Therefore (5.7)*=1055.6.

To find the third power of 0.12 log.

4

log. 1055.6 3.02348

0.75587

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 too 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 log. 296.35 = Therefore ✔87825 = 296.35.

=

4.94362

2.47181

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.

Therefore 30.57894

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. 398 = 2.59988 log. 4.893 = 0.68958 log. 58.729 = 1.76885

Therefore x=58.729.

To find the fourth term of the proportion

Therefore x=614.77.

0.0059768: 478.3 :: 0.0076821 : x,

(ar. co.) log. 0.0059768:
= 12.22353
log. 478.3 = 2.67970
log. 0.0076821 7.88548
2.78871

log. 614.77 =

=

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

To find the value of the fraction

1.

957

2.98091

0.99364

410 = 2 2.61278

5.22556

8.78932

log. 2 log. log. 0.00379= | | 17.57864 log. 4.653 = | 0.66773=0.13355= (ar. co.) 9.86645 4 log. 2.39 = 4 0.37840=1.51360= (ar. co.) 8.48640 log. 2298.1 3.36137 Therefore the value of the fraction is 2298.1.

2.

3.

=

EXAMPLE.

5.

957×(410)3×(0.00379)
(4.653)*X(2.39)*

82. The following examples will afford practice in all the usual operations by logarithms.

(0.00015)3×(0.0000156)5__5.5747

(587.69)

=

(10)35

4. √(149.16)3 —(17.8)2=148.09. (See Ex. 7, Art. 67.)

(567.8)-(467.2)
(15)7×0.53

= = 0.0011498.

ax=b,

6. √(49.755)2+(37.845)3=62.513. (See Ex. 8, Art. 67.)

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

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

0.90309

whence we have x

log. 50

log. 8

=

we have

=

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

log. 1.69897 0.23019
log. 0.903099.95573

log. 1.8813=0.27446

84. The complete solution of an exponential equation is therefore as follows:

Taking the logarithms of the equation

a∞ = b,

x log. a = log. b,

log. b

log. a'

whence

and again taking the logarithms

log.x=

log. b log.

log. a

x=

=

1.8813.

EXAMPLE

log. 100=2.00000,
log. 121.07918,

= log. log. b—log. log. a.

To find the value of x in the equation

12100.

= 0.30103

log. 2.00000
log. 1.07918=0.03309

log. 1.8533=0.26794

Therefore x = = 1.8533, or 121.9533— 100; that is, 1.8533 is the lo garithm of 100 in the system whose base is 12.