Logarithms
Subject :
Mathematics
Topic :
Logarithms
Class :
SS 1
Term :
First Term
Week :
Week 6
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Previous Knowledge :
The pupils have previous knowledge of
Standard form and Indices
Behavioural Objectives: At the end of the lesson, the students should be able to
 define logarithm
 deduce a relationship between logarithm and indices
 use the graph of y=10x for multiplication and division
 find the logarithm and antilog of numbers greater than one
Content:
TOPIC: LOGARITHMS CONTENT
 Deducing logarithm from indices and standard form
 Definition of Logarithms
 Antilogarithms
 The graph of y = 10^{x}
There is a close link between indices and logarithms.
Similarly, “log base e of 2” can be written as “e to the power of 1/2”, since 2 = e1/2.
One of the most useful properties of logarithms is that they can be used to solve equations in which the unknown quantity is an exponent. For instance, suppose we want to find out if a number raised to the power of 3 equals 27. We can use logs to solve this equation.
3x = 27
Take the log from both sides:
log 3x = log 27
Now we can use the fact that logs obey the same laws as indices to simplify this equation:
x log 3 = log 27.
Divide both sides by log 3:
x = (log 27) / (log 3).
Now we just need to find the value of (log 27) / (log 3). This can be done using a calculator or by looking up the values in a table of logarithms. We find that
x = 3
So the solution to the equation 3x = 27 is x = 3.
This method can be used to solve any equation in which the unknown quantity is an exponent. All we need to do is take the log of both sides of the equation and then use the laws of indices to simplify it. Finally, we just need to find the value of the expression on the left hand side using a calculator.
100 = 10^{2}. This can be written in logarithmic notation as log_{10}100 = 2. Similarly 8 = 2^{3} and it can be written as log_{2}8 = 3.
In general, N = b^{x} in logarithmic notation is Log_{b}N = x.
We say the logarithms of N in base b is x. When the base is ten, the logarithms is known as common logarithms.
The logarithms of a number N in base b is the power to which b must be raised to get N.
Evaluation.
Rewrite using logarithmic notation (i) 1000 = 10^{3} (ii)0.01 = 10^{2} (iii)= 16 (iv) = 2^{3} Change the following to index form
(i) Log_{4}16 = 2 (ii) log_{3} () = 3
The logarithm of a number has two parts and integer (whole number) then the decimal point. The integral part is called the characteristics and the decimal part is called mantissa.
To find the logarithms of 27.5 form the table, express the number in the standard form as 27.5
= 2.75 x 10^{1}. The power of ten in this standard form is the characteristics of Log 27.5. The decimal part is called mantissa.
Remember a number is in the standard form if written as A x 10^{n} where A is a number such that 1 ≤ A < 10 and n is an integer.
27.5 = 2.75 x 10^{1,} 27.5 when written in the standard form, the power of ten is 1. Hence the characteristic of Log 27.5 is 1. The mantissa can be read from 4figure table. This 4figure table is at the back of your New General Mathematics textbook.
Below is a row from the 4figure table

Differences

To check for Log27.5, look for the first two digits i.e 27 in the first column.
Now look across that row of 27 and stop at the column with 5 at the top. This gives the figure 4393.
Hence Log27.5 = 1.4393
To find Log275.2, 2.752 x 10^{2}
The power of 10 is the standard form of the number is 2. Thus, the characteristic is 2. Log275.2
= 2. ‘Something’
For the mantissa, find the figure along the row of 27 with middle column under 5 as before (4393). Now find the number in the differences column headed. This number is 3. Add 3 to 4393 to get 4396. Thus Log275.2 = 2.4396.
Evaluation
Use table to find
 Log 1
 Log 71
 Log 238
LESSON 2
If the logarithm of a number is given, one can determine the number for the antilogarithm table.
Example: Find the antilogarithms of the following (a) 0.5670 (b)2.9504 Solution
 The first two digits after the decimal point i.e .56 is sought for in the extreme left column of the antilogarithm table then look across towards right till you, get to the column with heading 9. (Read 56 under 9) there you will see 3707. Since the integral part of 0.5690 is 0, it means if the antilog (3707) is written in the standard form the power of 10 is
i.e antilog of 0.5690 = 3.707 x 10^{0}
= 3.707
 The antilog of 9504 is found by checking the decimal part .9504 in the table.
 Along the row beginning with 0.95 look forward right and pick the number in the column with 0 at the top. This gives the figures 8913 proceed further to the difference column with heading 4 to get This 8 is added to 8913 to get 8921. The integral part of the initial number (2.9504) is
 This shows that there are three digits before the decimal point in the antilog of 2.9504 so
antilog 2.9504 = 892.1.
LESSON 3
The graph of y = 10^{x} can be used to find antilogarithm (and logarithm). Below is the table of values.
X  0  0.1  0.2  0.3  0.4  0.5  0.6  0.7  0.8  0.9  1 
Y = 10^{x}  1  1.3  1.6  2.0  2.5  3.2  4.0  5.0  6.3  7.9  10 
For example the broken line shows that the antilog 0.5 is approximately 3.2 or inversely that Log 3.2 ~0.5
Evaluation
 Find the logarithms of
 (i) 32.7
 (ii) 61.02
 (iii) 3.247
 Use antilog tables to find the numbers whose logarithms are
 (1) (1.82)
 (ii)2.0813
 (iii) 2108
Natural Log Rules
There are a large number of logarithmic rules. However, the major ones are listed below:
 Logarithm product rule: logarithm of the multiplication of x and y is the sum of the logarithm of x and logarithm of y. [log_{b}(x ∙ y) = log_{b}(x)+ log_{b}(y)]
For example: log_{10}(4 ∙ 8) = log_{10}(4) + log_{10}(8)
 Logarithm quotient rule: logarithm of the division of x and y provides us with the difference of logarithm of x and logarithm of y. [log_{b}(x / y) = log_{b}(x) log_{b}(y)]
For example: log_{10}(4 / 8) = log_{10}(4) – log_{10}(8)
 Logarithm power rule: logarithm of x raised to the power of y gives us y times the logarithm of x. [log_{b}(x ^{y}) = y ∙ log_{b}(x)]
For example: log_{10}(3^{8}) = 8∙ log_{10}(3)
 Logarithm base switch rule: logarithm of x to the base y is equal to 1 divided by logarithm of y to the base x. [log_{b}(x) = 1 / log_{c}(y)]
For example: log_{3}(4) = 1 / log_{4}(3)
 Logarithm base change rule: logarithm of x to the base y is equal to the logarithm of x to the base z divided by the logarithm of y to the base z minus logarithm of x. [log_{y}(x) = log_{z}(x) / log_{c}(b) Logarithm – log(x)]
For example: log_{4}(3) = log_{5}(3) / log_{5}(4) Logarithm – log(3)
Logarithms II
LOGARITHMS
CONTENT:
 Use Of Logarithm Table And Antilogarithm Table In Calculations Involving
 Multiplication
 Division
 Powers
 Roots
To multiply two numbers, look up each number in the logarithm table and add the logs together. Then, find this sum in the antilogarithm table to get the product
For example, to calculate 72 9 = 648, we would look up 7 and 2 in the logarithm table
Add the logs together to get 2.879 (which rounds to 2.88 in the antilogarithm table)
Then look up 2.88 in the antilogarithm table to find that 648 is the product
2. To divide two numbers, look up each number in the logarithm table and subtract the logs. Then, find this difference in the antilogarithm table to get the quotient
For example, to calculate 72 / 9 = 8, we would look up 7 and 2 in the logarithm table
Subtract the logs to get 2.879 (which rounds to 2.88 in the antilogarithm table)
Then look up 2.88 in the antilogarithm table to find that 8 is the quotient
3. To calculate a power, look up the base in the logarithm table and multiply by the exponent. Then, find this product in the antilogarithm table
For example, to calculate 72 = 4, we would look up 7 in the logarithm table and multiply by 2
This gives us 2.879 (which rounds to 2.88 in the antilogarithm table)
Then look up 2.88 in the antilogarithm table to find that 4 is the answer
4. To calculate a root, look up the number whose root you want to take in the logarithm table and divide by the exponent. Then, find this quotient in the antilogarithm table
For example, to calculate 72 = 9, we would look up 7 in the logarithm table and divide by 2
This gives us 2.879 (which rounds to 2.88 in the antilogarithm table)
Then look up 2.88 in the antilogarithm table to find that 9 is the answer
5. You can also use the logarithm table and antilogarithm table to calculate exponents
For example, to calculate e2 = 7.389, we would look up 2 in the logarithm table
This gives us 0.693 (which rounds to 0.69 in the antilogarithm table)
Then look up 0.69 in the antilogarithm table to find that 7.389 is the answer
In summary, the logarithm table and antilogarithm table can be used to simplify calculations involving multiplication, division, powers, and roots. These tables are also useful for calculating exponents.
Multiplication of Numbers using Logarithm Tables
Multiplication is one of the four basic operations in mathematics, along with addition, subtraction, and division. Although it might seem simple to multiply two numbers, there are actually a number of different ways to do so, each with its own advantages and disadvantages. One method is to use logarithm tables.
Logarithm tables are tables of numbers that have been calculated to logarithmic form. They can be used to simplify multiplication by allowing you to add the logarithms of the numbers you’re multiplying instead of multiplying the numbers themselves. For example, if you wanted to multiply 2 times 3, you could look up the logarithm of 2 in the table and add it to the logarithm of 3. This would give you the logarithm of 6, which you could then convert back into a regular number to get your answer.
Logarithm tables are most often used by people who need to do a lot of multiplication or who are working with very large numbers. They can be found in mathematical handbooks and are sometimes used by engineers and scientists.
1. To multiply 2 times 3, look up the logarithm of 2 in the table and add it to the logarithm of 3. This gives you the logarithm of 6, which you can convert back into a regular number to get your answer.
2. To multiply 3 times 4, look up the logarithm of 3 in the table and add it to the logarithm of 4. This gives you the logarithm of 12, which you can convert back into a regular number to get your answer.
3. To multiply 4 times 5, look up the logarithm of 4 in the table and add it to the logarithm of 5. This gives you the logarithm of 20, which you can convert back into a regular number to get your answer.
4. To multiply 5 times 6, look up the logarithm of 5 in the table and add it to the logarithm of 6. This gives you the logarithm of 30, which you can convert back into a regular number to get your answer.
5. To multiply 6 times 7, look up the logarithm of 6 in the table and add it to the logarithm of 7. This gives you the logarithm of 42, which you can convert back into a regular number to get your answer.
LOGARITHMS
CONTENT: Use Of Logarithm Table And Antilogarithm Table In Calculation Involving
 Multiplication
 Division
 Powers
 Roots
 Application of logarithm in capital market and other real life problems
PERIOD 1
Logarithm and antilogarithm tables are used to perform some arithmetic basic operations namely: multiplication and division. Also, we use logarithm in calculations involving powers and roots.
The basic principles of calculation using logarithm depends strictly on the laws on indices. Recall from week 7 that.
 Log MN = Log M + Log N
 Log = Log M – Log N
Hence, we conclude that in logarithm;
 When numbers are multiplied, we add their logarithms
 When two numbers are dividing, we subtract their
SUB TOPIC 1
Multiplication of numbers using logarithm tables
Example 1
Evaluate 92.63 x 2.914 Solution
Therefore 92.63 x 2.914 = 269.9
Example 2
Evaluate 34.83 x 5.427 Solution
Therefore 34.83 x 5.427= 189.1
EVALUATION
Evaluate the following 1. 6.26 x 23.83
 409.1 x 3.932
 8.31 x 22.45 x 19.64
 431.2 x 21.35
SUBTOPIC 2 – DIVISION OF NUMBERS USING LOGARITHM
Example 1
Evaluate 357.2 ÷ 87.23
Solution1
Therefore 357.2 x 87.23 = 4.096
Example 2
Use a logarithm table to evaluate 75.26 ÷ 2.581
Solution
Number 
Standard Form 
Log 
Operation 
75.26  7.526 x 10^{1}  1.8765  Subtraction 
2.581  2.581 x 10^{0}  0.4118  
1.4647 
Log AntiLog Standard form Number
1.4647 2916 2.916 x 10^{1} 29.16
Therefore 75.26 ÷ 2.581 = 29.16
Evaluation
Use table to evaluate the following
(1) 53.81 ÷ 16.25 (2)632.4 ÷ 34.25 (3) 63.75 ÷ 8.946 (4) 875.2 ÷ 35.81
SUBTOPIC 3
Calculation Of Powers Using Logarithm
Study these examples.
At times, calculations can involve powers and roots. From the laws of logarithm, we have
 Log M^{n} = nLogM
 Log M^{1/n} = Log M = Log
 Log M^{x/n} = Log M =
Example 1.
Evaluate the following (53.75)^{3}
Solution
Therefore 53.75^{3} = 155300
Example 2: 64.59^{2}
Solution
Therefore 64.59^{2}= 4173
Evaluation
Evaluate the following 1. 5.632^{4}
2.
 19.18^{3}
 (67.9/5.23)^{3}
 x 92.6
SUBTOPIC 4
Calculation of roots using Logarithms and AntiLogarithms
Use tables of Logarithms and antilog to calculate Solution
Therefore = 1.939
Example 2: use table to find Solution
Hint: Workout 218 ÷ 3.12 before taking the cube root.
Number  Standard Form  Log  Operation  
218  (2.18 x 10^{2})  2.3385  Subtraction  
3.12  (3.12 x 10^{0})  0.4942  
1.8  443 ÷ 3 Divisi
0.6148 
on 
Therefore = 4.119
Example 3
Evaluate 63.75^{2} – 21.39^{2}
We can use difference of two squares
i.e A^{2} – B^{2} = (A+B)(AB)
63.75^{2} – 21.39^{2} = (63.75+21.39)(63.7521.39)
= (85.14)(42.36)
= 85.14 X 42.36
Number  Log  operation 
85.14  (1.9301)  Addition 
42.36  (1.6260)  
3606  3.5561  
…. 3606  
EVALUATION 
Evaluate the following
 (39.65^{2} – 7.43^{2})^{1/2}
 84.35^{2} – 36.95^{2}
 64.74^{2} – 55.262
 94.68^{2} – 43.25^{2}
 25.14^{2} – 7.52^{2}
Presentation
The topic is presented step by step
Step 1:
The subject teacher revises the previous topics
Step 2.
He or she introduces the new topic.
Step 3:
The subject teacher allows the pupils to give their own examples and he corrects them when the need arises.
Conclusion:
The subject teacher wraps up or concludes the lesson by giving out a short note to summarize the topic that he or she has just taught.
The subject teacher also goes round to make sure that the notes are well copied or well written by the pupils.
He or she makes the necessary corrections when and where the needs arise.