Indices and Logarithms  Mathematics Form 2
LESSON OBJECTIVES
By the end of the lesson the learner should be able to: perform the basic operations on bar numbers accurately.
Logarithms are applied in almost every part of life directly and indirectly.
They help us to simply work involving numbers. Discover how this is applicable!
LESSON INTRODUCTION
Expressing numbers as powers of 10
Numbers can be written such that 10 is the base. Power of 10 is called a logarithm.
E.g. express the following numbers as powers of 10
1000 = 103
100 = 102
10 = 101
1 = 100
0.1 = 10 1
0.01=102
0.001=103
The powers 3,2,1,0,1,2 and 3 of the numbers 1000, 100, 10, 1, 0.1, 0.01 and 0.001 respectively are known as their logarithms.
The logarithm of 100, written as log 100 = 2 since 100 = 102
The logarithm of a number can be read form logarithm table.
A part of mathematical table
e.g Logarithm of 4.926 is obtained as follows:
In the column headed x read 4.9. then proceed horizontally to the column headed 2 read 0.6920. (Animate the movement of the arrow for 1.0 to 4.9 and them to 0.6920).
In the mean difference columns proceed. Further horizontally and read 5 in the column headed 6.
To get the logarithm add 5 to 0.6920 as shown below:
0.6920
5+
0.6925
Therefore, Log 4.926 = 0.6925
Other examples:
Example 1
Find Log 6.72
Solution
Write the number in standard form i.e. 6.72 = 6.72 x 10o
From the logarithm tables the log of 6.72 = 0.8274
Example 2
 Find log 32.3
Solution
Write 32.3 in standard form
32.3 =3.23 x 101
Log 32.3 = 1.5092
Note
The logarithm of a number has two parts: the integral part which is called the characteristic and the decimal part which is called the mantissa. Above6.72 was written as 6.72x106
Example 1
Log 6.72 =0.8274
From example 2, 32.3 is written as 3,23 x 101
log 32.3 = 1.5092
Note:
The powe of 10 when a number is written in standard form is the characteristic.
(Highlight in different colours both he characteristic and mantissa. The colour of the power of 10 should match with that of the characteristic. )
Example 3
 Find log 0.0079
Solution
Write 0.0079 in standard form
0.0079 = 7.9 x 103
In this example the power of 10 is 3. The characteristic is written as 3 with the negative sign on top i.e. and read as bar 3. Therefore the logarithm of 0.0079=
For numbers that i.e between 0 and 1 the characteristic is always negative and the mantissa is always positive.
Examples 4
Evaluate the following using logarithms
728 x 32.6
Solution
When multiplying numbers using logarithms write them in standard form and read the logarithms from mathematical tables. Arrange them as shown below
The logarithms are added to get 4.3753 from the antilogarithm tables.
Note finding antilogarithm is the revenue of finding logarithm.
To find the expected product then read the antilogarithm of 4.3753 from the antilogarithm tables.
Note: Finding antilogarithm is the revenue of finding logarithm.
Using antilogarithm tables read the antilog of 0.3753 (the mantissa)
This gives 2.373
The characteristic 4 in 4.3753 above is simply the power of 10 when the product is written in standard form.
Therefore the expected product is
2.373 x 104 = 23730.
Note:
To multiply two numbers we add the logarithms.
Example 5
Evaluate 48.12 6.43 using logarithm tables.
Solution
Arrange the walking as
Number 
Standard form 
log 
48.12 
4.812 x 101 6.43 x 10o 
1.6823  0.88082 
0.7741 
To divide the two numbers we subtract their logarithms
Then find the antilogarithm of 0.7741 which is 5.944.
Note that the characteristic is 0. Therefore the answer becomes
= 5.944 x 10o = 5.944 x 1
= 5.944
Using logarithm tables
The working is arranged as follows:
NUMBER 
STANDARD FORM 
FORM 
2.61 
2.61 x 10o 
0.4166 
61.72 
6.172 x 101 
1.7904 
21.8 
2.18 x 01 
2.2070 
In finding the cube rout of a number is simply multiplying the number with a 1/3 or divide it by 3
In this case 0.8685 is then divided by 3 i.e
Then find the antilog of 0.2895
Which is 1.948
Note
To find the nth root of a number in an expression We divide the log of b by n.
Lesson objectives
By the end of the lesson, you should you able to perform basic operations on logarithm numbers whose characteristic in negative accurately.
Lesson development
In this lesson you are going to learn about the basic operations on logarithm numbers whose characteristic in negative .
Logarithms of numbers between 0 and 1
Logarithm of numbers between 0 and 1 have a negative characteristic but the mantissa is always positive.
e.g
Log 0.02 = log (2.0 x 102)
=
Such logarithms are known as bar numbers can be written as 2 + 0.3010
Therefore bar is simply a minus sign written above 2 as and read as bar 2.
We are now going to perform basic operations on bar numbers.
Addition of logarithm numbers whose characteristic in negative
Example 1
Evaluate
This is rewritten and arranged as:
1 + 0.1502 +
2 + 0.1051
3 + 0.2553
This gives
Example 2
Evaluate
Solution
Rewrite and arrange as
3 +0.4213 +
1 + 0.7055 (Note: 4+1=3)
4 + 1.1268 = 3 + 0.1268
= 3.1268
Subtraction of logarithm numbers whose characteristic in negative
Example 1
Evaluate
Written as  4+ 0.5082 
2 + 0.4701
2+0.0381 =
Example 2
Evaluate
Solution
1+0.6301 
3+0.8015
It is not possible to subtract 0.8015 from 0.6301, therefore we borrow (substract) 1 from 1(characteristic i.e. 1 1=2 and then proceed as follows
2+1.6301
3+0.8015
1 0.8286
(Note that 2 3 = 2+3 = 3 2 =1)
Example 3
(iii)
1+0.3701 
3+0.8123
We borrow 1 from 1.3701 and rewrite it as 0+ 1.3701
2+0.8123
2 + 0.5578 = 2.5578
Multiplication of logarithm numbers whose characteristic in negative
Example 1
Evaluate
solution
Rearrange as 2+0.3501
x 3
 6+1.0503 = 5+0.0503
=
Example 2
Evaluate:
solution
Arrange as 2+0.1231
x 2
4 + 0.246
Note:
In logarithms the mantissa should be always be positive. In case it is negative, it is changes to be positive as shown below. Borrow 1 form 4 targer and then proceed as follows:
3 + 1 0.2462 =
This becomes 3 + 0.7538 = 3.7538. which is equal to
(d) Division of logarithm numbers whose characteristic in negative
Example 1
Evaluate
Solution
Rewrite the number as 2+0.1608
Then divide each by 2 as shown below:
Example 2
Evaluate
The characteristic should be a whole number that is divisible by 2. We think of a possible way of making it divisible by 2. e.g substracy 1 form the characteristic (1) and add 1 to the mantissa i.e. 1 1=2 and 1 + 0.560=1.5060. This gives
2 + 1.5060. Divide (2+1.5060 by 2 as follows:
Logarithm Computations of numbers whose logarithms have negative characteristics
Example 1
Work out 0.0256 x 0.5792
Solution
Arrange the working as
Number 
standard form 
log 
0.0256 
2.56 x 102 

0.5796 
5.5796 x 10 1 



Find antilog of
=1.4829 x102 = 0.01429
Example 2
Evaluate using logarithms
Solution.
is written as
Borrow 2 form 1 to make the characterisctic divisible by 3 and add 2 to the mantissa 0.1898
=
Find the antilog of =5.36911=0.53691
Example 3
0.3263 x 42.27
Solution.
Number 
Standard Form 
Log 
0.3263 
3.26 x 101 

Find antilog of 0.1656
= 1.4642 x 10o = 1.4642 x 1
= 1.4642.
Summary
 An expression such as can be written as a 1/ne.g. , square root sign means raising a number to power , similarly
 When multiplying numbers simply add their logarithms and dividing numbers subtract their logarithms
 The characteristic in a logarithm is always the power into which 10 is raised when a number is written in standard form.
Lesson Objectives
By the end of the lesson, you should you able to perform basic
operations on bar numbers accurately.
Expressing numbers as powers of 10
Numbers can be written such that 10 is the base. Power of 10 is called a logarithm.
v/o For example express the following numbers as powers of 10
1000 = 103
100 = 102
10 = 101
1 = 100
0.1 == 10 1
0.01=
0.001=
The powers 3,2,1,0,1,2 and 3 of the numbers 1000, 100, 10, 1, 0.1, 0.01 and 0.001 respectively are known as their logarithms.
The logarithm of 100, written as log 100 = 2 since 100 = 102
The logarithm of a number can be read form logarithm table.
A part of mathematical table


Add Mean differences 
x 
0 1 2 3 4 5 6 7 8 9 
1 2 3 4 5 6 7 8 9 
1.0 4.9 
0.6920 
5 
e.g. Logarithm of 4.926 is obtained as follows:
In the column headed x read 4.9. then proceed horizontally to the column headed 2 read 0.6920. (Animate the movement of the arrow for 1.0 to 4.9 and them to 0.6920).
In the mean difference columns proceed. Further horizontally and read 5 in the column headed 6.
To get the logarithm add 5 to 0.6920 as shown below:
0.6920
5+
0.6925
Therefore, Log 4.926 = 0.6925
v/o Below are other examples
Example 1
Find Log 6.72
Solution
Write the number in standard form i.e. 6.72 = 6.72 x 10o
From the logarithm tables the log of 6.72 = 0.8274
Example 2
 Find log 32.3
Solution
Write 32.3 in standard form
32.3 =3.23 x 101
Log 32.3 = 1.5092
Note
The logarithm of a number has two parts: the integral part which is called the characteristic and the decimal part which is called the mantissa.
In example 1 above 6.72 was written as 6.72x100
v/o therefore, the
Log 6.72 =0.8274
From example 2, 32.3 as written as 3.23 x 101
v/o therefore, the
log 32.3 = 1.5092
Note:
The power of 10 when a number is written in standard form is the characteristic.
(Highlight in different colours both he characteristic and mantissa. The colour of the power of 10 should match with that of the characteristic. )
Example 3
Find log 0.0079
Solution
Write 0.0079 in standard form
0.0079 = 7.9 x 103
In this example the power of 10 is 3. The characteristic is written as 3 with the negative sign on top i.e. and read as bar 3. Therefore the logarithm of 0.0079=
For numbers that i.e. between 0 and 1 the characteristic is always negative and the mantissa is always positive.
Roots of numbers using logarithms
Example
Evaluate
Using logarithm tables
Solution
The working is arranged as follows:
NUMBER 
STANDARD FORM 
FORM 
2.61 
2.61 x 10o 
0.4166 
61.72 
6.172 x 101 
1.7904 
21.8 
2.18 x 101 
2.2070 
In finding the cube root of a number is simply multiplying the number with a 1/3 or divide it by 3
0.8685 is then divided by 3 i.e. 0.8685 = 0.2895
3
Then the antilog of 0.2895
= 1.948 x 10o = 1.948
NB. To find the n the root of a number in an expression
n 6 We divide the log of b by n.
Logarithm of numbers between 0 and 1
Logarithm of numbers between 0 and 1 have a negative characteristic but the mantissa is always positive.
e.g.
Log 0.02 = log (2.0 x 102)
= 2.3010
Such logarithms are known as bar numbers 2.3010 can be written as 2 + 0.3010
Therefore bar is simply a minus sign written on 2 as 2 and read as bar 2 as started earlier.
Basic operations on logarithm numbers with negative characteristics.
(a) Addition
Example 1
Evaluate
This is rewritten and arranged as:
1 + 0.1502 +
2 + 0.1051
3 + 0.2553
This gives
Example 2
Evaluate
Solution
Rewrite and arrange as
3 +0.4213 +
1 + 0.7055 (Note: 4+1=3)
4 + 1.1268 = 3 + 0.1268
= 3.1268
(b) Subtraction
Example 1
Evaluate
Written as  4+ 0.5082 
2 + 0.4701
2+0.0381 =
Example 2
Evaluate
Solution
1+0.6301 
3+0.8015
It is not possible to subtract 0.8015 from 0.6301, therefore we borrow (substract) 1 from 1(characteristic i.e. 1 1=2 and then proceed as follows
2+1.6301
3+0.8015
1 0.8286
(Note that 2 3 = 2+3 = 3 2 =1)
Example 3
(iii)
1+0.3701 
3+0.8123
We borrow 1 from 1.3701 and rewrite it as 0+ 1.3701
2+0.8123
2 + 0.5578 = 2.5578
Multiplication
Example 1
Evaluate
Rearrange as 2+0.3501
x 3
 6+1.0503 = 5+0.0503
=
Example 2
Evaluate:
Arrange as 2+0.1231
x 2
4 + 0.246
Note:
In logarithms the mantissa should be always be positive. In case it is negative, it is changes to be positive as shown below. Borrow 1 form 4 target and then proceed as follows:
3 + 1 0.2462 =
This becomes 3 + 0.7538 = 3.7538. which is equal to
(d) Division
Example 1
Evaluate
Solution
Rewrite the number as 2+0.1608
Then divide each by 2 as shown below:
Example 2
Evaluate
The characteristic should be a whole number that is divisible by 2. We think of a possible way of making it divisible by 2. e.g substracy 1 form the characteristic (1) and add 1 to the mantissa i.e. 1 1=2 and 1 + 0.560=1.5060. This gives
2 + 1.5060. Divide (2+1.5060 by 2 as follows:
Addition of bar numbers
Example 1
Evaluate
This is rewritten and arranged as:
1 + 0.1502 +
2 + 0.1051
3 + 0.2553
This gives
Example 2
Evaluate
Solution
Rewrite and arrange as
3 +0.4213 +
1 + 0.7055 (Note: 4+1=3)
4 + 1.1268 = 3 + 0.1268
= 3.1268
Subtraction of bar numbers
Example 1
Evaluate
Written as  4+ 0.5082 
2 + 0.4701
2+0.0381 =
Example 2
Evaluate
Solution
1+0.6301 
3+0.8015
It is not possible to subtract 0.8015 from 0.6301, therefore we borrow (substract) 1 from 1(characteristic i.e. 1 1=2 and then proceed as follows
2+1.6301
3+0.8015
1 0.8286
(Note that 2 3 = 2+3 = 3 2 =1)
Example 3
(iii)
1+0.3701 
3+0.8123
We borrow 1 from 1.3701 and rewrite it as 0+ 1.3701
2+0.8123
2 + 0.5578 = 2.5578
Multiplication of bar numbers
Example 1
Evaluate
Rearrange as 2+0.3501
x 3
 6+1.0503 = 5+0.0503
=
Example 2
Evaluate:
Arrange as 2+0.1231
x 2
4 + 0.246
Note:
In logarithms the mantissa should be always be positive. In case it is negative, it is changes to be positive as shown below. Borrow 1 form 4 targer and then proceed as follows:
3 + 1 0.2462 =
This becomes 3 + 0.7538 = 3.7538. which is equal to
Division of bar numbers
Example 1
Evaluate
Solution
Rewrite the number as 2+0.1608
Then divide each by 2 as shown below:
Example 2
Evaluate
The characteristic should be a whole number that is divisible by 2. We think of a possible way of making it divisible by 2. e.g substracy 1 form the characteristic (1) and add 1 to the mantissa i.e. 1 1=2 and 1 + 0.560=1.5060. This gives
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