Indices and Logarithms | Mathematics Form 2

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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=10-2

0.001=10-3

 


    

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

  1. 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

  1. Find log 0.0079

Solution
Write 0.0079 in standard form
9      0.0079 = 7.9 x 10-3
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
6.43


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
13 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
-1.3385
 0.8685

             
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  14        

Then find the antilog of 0.2895
Which is 1.948
Note
To find the nth root of a number in an expression 15 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 10-2)
= 16
Such logarithms are known as bar numbers 17 can be written as -2 + 0.3010
Therefore bar is simply a minus sign written above 2 as 18 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 19
This is re-written and arranged as:
-1 + 0.1502   +
-2 + 0.1051
-3 + 0.2553
This gives 20
Example 2
Evaluate 21

Solution
Rewrite and arrange as
-3 +0.4213   +
-1 + 0.7055    (Note: -4+1=-3)
      -4 + 1.1268 = -3 + 0.1268
22      = 3.1268

Subtraction of logarithm numbers whose characteristic in negative

      Example 1

Evaluate 23
      Written as - 4+ 0.5082 -
           -2 + 0.4701
          -2+0.0381 = 24
Example 2
Evaluate 24
      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)    26     
            -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 27

solution
Re-arrange as -2+0.3501
28            x 3
    - 6+1.0503 = -5+0.0503
=29 

Example 2
Evaluate: 30

solution

Arrange as   -2+0.1231
         x -2
31      32
        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 33
Solution
Rewrite the number as -2+0.1608
   Then divide each by 2 as shown below:
   34
    
   Example 2
   Evaluate 35
     36
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 10-2

37

0.5796


5.5796 x 10 -1

38



39


              
     
      Find antilog of  40
      =1.4829 x10-2 = 0.01429
Example 2
Evaluate using logarithms
41
        
Solution.
42 is written as 43
Borrow 2 form -1 to make the characterisctic divisible by 3 and add 2 to the mantissa 0.1898
44
=45
Find the antilog of 46=5.3691-1=0.53691
Example 3
0.3263 x 42.27
Solution.


Number

Standard Form

Log

0.3263


3.26 x 10-1

47
     x3
48

      

          
                    
Find antilog of 0.1656
            = 1.4642 x 10o = 1.4642 x 1
            = 1.4642.
Summary

  1.  An expression such as  49can be written as    a 1/ne.g.   50, square root sign means raising a number to power , similarly    51  
  2. When multiplying numbers simply add their logarithms and dividing numbers subtract their logarithms
  3. 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 =1= 10 -1
           
        
0.01= 1
       
         0.001=1   

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

10  1  2  3  4  5  6  7  8  9

11  2  3  4  5  6  7  8  9

11.0


4.9


11  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
1          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

  1. 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
1      0.0079 = 7.9 x 10-3
In this example the power  of 10 is -3. The characteristic is written as 3 with the negative sign on top i.e. 1 and read as bar 3. Therefore the logarithm of 0.0079=1

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
3 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
-1.3385
 0.8685

             
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
333NB. 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 10-2)
4= 2.3010
4Such 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 re-written 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
Re-arrange 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 1
This is re-written and arranged as:
-1 + 0.1502   +
-2 + 0.1051
-3 + 0.2553
This gives 1
Example 2
Evaluate 1

Solution
Rewrite and arrange as
-3 +0.4213   +
-1 + 0.7055    (Note: -4+1=-3)
      -4 + 1.1268 = -3 + 0.1268
1      = 3.1268

Subtraction of bar numbers

  Example 1

Evaluate 12
      Written as - 4+ 0.5082 -
           -2 + 0.4701
          -2+0.0381 = 12
Example 2
Evaluate 12
      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)    12     
            -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 1
Re-arrange as -2+0.3501
1            x 3
    - 6+1.0503 = -5+0.0503
=1 

Example 2
Evaluate: 1
Arrange as   -2+0.1231
         x -2
1      1
        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 1
Solution
Rewrite the number as -2+0.1608
   Then divide each by 2 as shown below:
   2
  
   Example 2
   Evaluate 3
   4
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

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