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.)
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.
Get The Best Grades With the Least Amount of Effort
Here is a collection of proven tips, tools and techniques to turn you into a super-achiever - even if you've never thought of yourself as a "gifted" student.
The secrets will help you absorb, digest and remember large chunks of information quickly and easily so you get the best grades with the least amount of effort.
If you apply what you read from the above collection, you can achieve best grades without giving up your fun, such as TV, surfing the net, playing video games or going out with friends!
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.)
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.
Progressive Learning of Math : Antilog
Recently, I have found a series of math curricula (Both Hard Copy and Digital Copy) developed by a Lady Teacher who taught everyone from Pre-K students to doctoral students and who is a Ph.D. in Mathematics Education.
This series is very different and advantageous over many of the traditional books available. These give students tools that other books do not. Other books just give practice. These teach students “tricks” and new ways to think.
These build a student’s new knowledge of concepts from their existing knowledge. These provide many pages of practice that gradually increases in difficulty and provide constant review.
These also provide teachers and parents with lessons on how to work with the child on the concepts.
The series is low to reasonably priced and include
