Logarithms using Log table

Notes 10 Pages

Bengaluru City University

Bachelor of Business Administration

Contributed by

Prajwal Hallale

Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA
Logarithms using Log table
In mathematics, the logarithm table is used to find the value of the logarithmic
function. The simplest way to find the value of the given logarithmic function is by
using the log table.
Logarithmic Function Definition
The logarithmic function is defined as an inverse function to exponentiation. The
logarithmic function is stated as follows
For x, a > 0, and a≠1,
y= log
a
x, if x = a
y

Then the logarithmic function is written as:
f(x) = log
a
x
The most 2 common bases used in logarithmic functions are base e and base 10.
The log function with base 10 is called the common logarithmic function and it is
denoted by log
10
or simply log.
f(x) = log
10

The log function to the base e is called the natural logarithmic function and it is
denoted by log
e
.

f(x) = log
e
x
To find the logarithm of a number, we can use the logarithm table instead of using
mere calculation. Before finding the logarithm of a number, we should know about
the characteristics and mantissa part of a given number
• Characteristic Part – The whole part of a number is called the
characteristic part. The characteristic of any number greater than one is
positive, and if it is one less than the number of digits to the left of the
decimal point in a given number. If the number is less than one, the
characteristic is negative and is one more than the number of zeros to the
right of the decimal point.
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Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA

• Mantissa Part – The decimal part of the logarithm number is said to be the
mantissa part and it should always be a positive value. If the mantissa part is
in a negative value, then convert into the positive value.

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MESIOM Logarithms with table MTBDA

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Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA
How to Use the Log Table?
The procedure is given below to find the log value of a number using the log table.
First, you have to know how to use the log table. The log table is given for the
reference to find the values.
Step 1: Understand the concept of the logarithm. Each log table is only usable with
a certain base. The most common type of logarithm table is used is log base 10.
Step 2: Identify the characteristics and mantissa part of the given number.
For example, if you want to find the value of log
10
(15.27), first separate the
characteristic part and the mantissa part.
Characteristic Part = 15
Mantissa part = 27
Step 3: Use a common log table. Now, use row number 15 and check column
number 2 and write the corresponding value. So the value obtained is 1818.
Step 4: Use the logarithm table with a mean difference. Slide your finger in the
mean difference column number 7 and row number 15, and write down the
corresponding value as 20.

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Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA
Step 5: Add both the values obtained in step 3 and step 4.
That is 1818+20= 1838. Therefore, the value 1838 is the mantissa part.

Step 6: Find the characteristic part. Since the number lies between 10 and 100, (10
1

and 10
2
), the characteristic part should be 1.
Step 7: Finally combine both the characteristic part and the mantissa part, it
becomes 1.1838.

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Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA
2. Find the value of log
10
2.872
Solution:
Step 1: Characteristic Part= 2 and mantissa part= 872
Step 2: Check the row number 28 and column number 7.
So the value obtained is 4579.
Step 3: Check the mean difference value for row number 28 and mean difference
column 2. The value corresponding to the row and column is 3
Step 4: Add the values obtained in step 2 and 3, we get 4582. This is the mantissa
part.
Step 5: Since the number of digits to the left side of the decimal part is 1, the
characteristic part is less than 1. So the characteristic part is 0
Step 6: Finally combine the characteristic part and the mantissa part. So it
becomes 0.4582.
Therefore the value of log 2.872 is 0.4582.

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Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA
Sample :
Consider a number with four digits say 4839 and will observe the logarithm value
for this number with variations.
Sl no
Number
Characteristics
Mantissa
Log value
1
Log 4839
(4 digits – 1 ) = 3
6847
3.6847
2
Log 48392
(5 digits – 1 ) = 4
6847
4.6847
3
Log 483925
(6 digits – 1 ) = 5
6847
5.6847
4
Log 483.9
(3 digits – 1 ) = 2
6847
2.6847
5
Log 48.39
(2 digits – 1 ) = 1
6847
1.6847
6
Log 4.839
(1 digits – 1 ) = 0
6847
0.6847
7
Log 0.4839
- ( 0 zeros + 1 ) = - 1
6847
1.6847
8
Log 0.04839
- ( 1 zero + 1 ) = - 2
6847
2.6847
9
Log 0.004839
- ( 2 zeros + 1 ) = - 3
6847
3.6847
10
Log 0.00483925
- ( 2 zeros + 1 ) = - 3
6847
3.6847
11
Log 0.000483925
- ( 3 zeros + 1 ) = - 4
6847
4.6847
12
Log 483
(3 digits – 1 ) = 2
6839
2.6839
13
Log 48
(2 digits – 1 ) = 1
6812
1.6812
14
Log 4
(1 digits – 1 ) = 0
6021
0.6021
15
Log 0.4
- ( 0 zeros + 1 ) = - 1
6021
1.6021
16
Log 0.04
- ( 1 zero + 1 ) = - 2
6021
2.6021
17
Log 0.04000
- ( 1 zero + 1 ) = - 2
6021
2.6021
18
Log 0.048
- ( 1 zero + 1 ) = - 2
6812
2.6812

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Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA
Problems:
1. Find log 28.93

The given number is greater than 1. The number of digits before decimal
point is 2.
Therefore characteristics is 2 – 1 =1
Mantissa in log table for 28
th
row 9
th
column is 4609
Add : mean difference in the same row at 3
rd
column is 5
So fractional part is 4614

Therefore log 28.93 = 1.4614

2. Find log 0.08285

The given number is lesser than 1. The number of zeros after decimal point
is 1.
Therefore characteristics is - (1 + 1) = -2 i. e 2
Mantissa in log table for 82
nd
row 8
th
column is 9180
Add : mean difference in the same row at 5
th
column is 3
So fractional part is 9183

Therefore log 0.08285 = 2 . 9183

3. Find log 289.3

The given number is greater than 1. The number of digits before decimal
point is 3.
Therefore characteristics is 3 – 1 =2
Mantissa in log table for 28
th
row 9
th
column is 4609
Add : mean difference in the same row at 3
rd
column is 5
So fractional part is 4614

Therefore log 289.3 = 2.4614

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Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA
4. Simplify using log table: 83.67 * 0.0347 * 0.038

Solution : since log is not attached with the problem first prefix log word

Let x = 83.67 * 0.0347 * 0.038
Applying log on both sides

Log x = Log (83.67 * 0.0347 * 0.038)
Applying logarithm law
Log x = log 83.67 + Log 0.0347 + log 0.038
Note:
• The second term 0.0347 has 1 zero after decimal point therefore the
characteristic is negative 2 . The table value for mantissa should be searched
for 3470.
• The third term 0.038 has 1 zero after decimal point therefore the
characteristic is negative 2 . The table value for mantissa should be searched
for 3800.
Keep in mind in the above two cases although the characteristic is negative but
mantissa part is positive

Therefore Log x = 1.9226 + 2.5403 + 2.5798

Log x = 2 + 2 + 1.9226 +0.5403 + 0.5798

Log x = 4 + 3.0427
Log x = 1.0427
When log attached with a variable will shift from one side to other side it becomes
Antilog
Therefore x = antilog 1.0427
( search in antilog table page same as log table method)
x = 0.1104
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Smt DIVYASHREE D V,
MESIOM Logarithms with table MTBDA
5. Simplify using log table

22.384
0.0564
Let x = 22.384
0.0564

Taking log both sides

log x = log 22.384
0.0564

Log x = log 22.384 – log 0.0564
Log x = 1.3498 – 2.7513
( here in second term, mantissa is also negative)
Log x = - 2 +1.3498 – 0.7513
Log x = 2 + 0.5985
Log x = 2.5985

x = antilog of 2.5985
x = 396.8

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