# ANTILOG -- LOGARITHM TABLES, SOLVED EXAMPLES, EXERCISE

Logarithm Tables before Antilog

if you have not already done so.

Having the knowledge of Characteristic
and Mantissa is a prerequisite here.

## Examples in Antilog

NOTE: For simplicity, we may use 'log' in place
of 'logarithm' and 'logs' in place of 'logarithms'.

In the following problems, besides the knowledge of Characteristic and Mantissa, we make use of theAlgebra Formulas from Logarithms.

### Example 1 of Antilog

Solve the following problem in Logarithm Tables

Given log 2 = 0.3010, log 3 = 0.4771, find the number of digits in 312 x 28.

Solution to Example 1 of Logarithm Tables :

Let x = 312 x 28
Taking logarithms to base 10 on both sides, we get
log x = log (312 x 28)

We know, log of a product can be written as
the sum of the logs of the factors of the product
(See Formula 5 in Logarithms; Link given above.)

∴ log x = log (312) + log (28)

We know, in log of a power, the exponent multiplies the log.
(See Formula 7 in Logarithms; Link given above.)

∴ log x = 12 (log 3) + 8 (log 2) = 12(0.4771) + 8(0.3010)
= 5.7252 + 2.4080 = 8.1332
⇒ Characteristic of x = 8 ⇒ x has 9 digits.
Number of digits in 312 x 28 = 9. Ans.

### Example 2 of Antilog

Solve the following problem in Logarithm Tables

Find the number of zeros between the decimal point and
the first significant figure in the value of (0.0504)12
given that log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451.

Solution to Example 2 of Logarithm Tables :

Let x = (0.0504)12
x = {504 x 10(-4)}12 = (504)12 x 10(-4 x 12) = (504)12 x 10(-48)

504 is divisible by 9 [ Since 5 + 0 + 4 = 9 is divisible by 9].
∴ 504 = 9 x 56 = 9 x 7 x 8 = 32 x 7 x 23
(504)12 = (32 x 7 x 23)12= 32 x 12 x 712 x 23 x 12 = 324 x 712 x 236

x = 324 x 712 x 236 x 10(-48)
Taking logarithms to base 10 on both sides, we get
log x = log {3(24) x 7(12) x 2(36) x 10(-48) }

We know, log of a product can be written
as the sum of the logs of the factors of the product
(See Formula 5 in Logarithms; Link given above.)

log x = log {3(24)} + log {7(12)} + log {2(36)} + log {10(-48)}

We know, in log of a power, the exponent multiplies the log.
(See Formula 7 in Logarithms; Link given above.)

log x = (24) (log 3) + (12) (log 7) + (36) (log 2) + (-48) (log 10)
= (24)(0.4771) + (12)(0.8451) + (36)(0.3010) -48(1)
= 11.4504 + 10.1412 + 10.836 - 48 = 32.4276 - 48 = -15.5724

To make Mantissa positive, -15.5724 is written as -16 + 0.4276
Thus log x = 16.4276⇒ Characteristic of x = 16
x has 15 zeros between the decimal point and the first significant figure.

The number of zeros between the decimal point and
the first significant figure in the value of (0.0504)12 is 15. Ans.

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### Example 3 of Antilog

Solve the following problem in Logarithm Tables

If log10 2 = 0.3010, show that log5 64 = 2.584

Solution to Example 3 of Logarithm Tables :

Let x = log5 64

We know, log of a quantity to a base can be written as the ratio of log of the quantity and log of the base.
(See Formula 8 in Logarithms; Link given above.)

x = (log 64)⁄(log 5) = (log 26)⁄{log (10⁄2)}

Numerator is logarithm of a power:

We know, in log of a power, the exponent multiplies the log.
(See Formula 7 in Logarithms; Link given above.)

Denominator is logarithm of a quotient:

We know, log of a quotient can be written
as the difference of the logs of the numerator
and denominator of the quotient
(See Formula 6 in Logarithms; Link given above.)

x = {6 (log 2)}⁄{log 10 - log 2} = 6(0.3010)⁄(1 - 0.3010)
= (1.806)⁄(0.699) = 2.5837 ≈ 2.584 (proved.)

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### Exercise of Antilog

Solve the following problems in Logarithm Tables

1. Given log 2 = 0.3010, find the number of digits in 12831.
2. Find the number of zeros between the decimal point
and the first significant figure in the value of (0.0432)10
given that log 2 = 0.3010, log 3 = 0.4771
3. If log10 3 = 0.4771, show that log30 81 = 1.292

For Answers to the problems in Logarithm Tables,
see at the bottom of the page.

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### Answers to Exercise of Antilog

The Answers to the problems in Antilog,
are given below.

1. 66
2. 33
3. Proof