ANTILOG -- LOGARITHM TABLES, SOLVED EXAMPLES, EXERCISE
Please study
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.
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.)
Exercise of Antilog
Solve the following problems in Logarithm Tables
- Given log 2 = 0.3010, find the number of digits in 12831.
- 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 - 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.
Answers to Exercise of Antilog
The Answers to the problems in Antilog, are given below. - 66
- 33
- Proof


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