Indices and Logarithms
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
- 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 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
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
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 10-2)
=
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 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
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
Re-arrange 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 10-2 | |
0.5796 | 5.5796 x 10 -1 | |
Find antilog of
=1.4829 x10-2 = 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
=5.3691-1=0.53691
Example 3
0.3263 x 42.27
Solution.
Number |
Standard Form | Log |
0.3263 | 3.26 x 10-1 |
x3 |
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 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.
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 -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
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 10-2)
=
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 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
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
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
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
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
Indices and Logarithms
e-Content
Buy e-Content Digital CD covers all the topics for a particular class per year. One CDs costs 1200/-
click to play video
Purchase Online and have the CD sent to your nearest Parcel Service. Pay the amount to Patrick 0721806317 by M-PESA then provide your address for delivery of the Parcel.. Ask for clarification if in doubt,
Indices and Logarithms
Candidate benefit from our quick revision booklets which are comprehensive and how to tackle examination question methods
We have an enourmous data quiz bank of past papers ranging from 1995 - 2017
KCSE ONLINE WEBSITE provide KCSE, KCPE and MOCK Past Papers which play a great role in students� performance in the KCSE examination. KCSE mock past papers serves as a good motivation as well as revision material for the major exam the Kenya certificate of secondary education (KCSE). Choosing the KCSE mock examination revision material saves you a lot of time spent during revision for KCSE . Choosing the KCSE mock examination revision material saves you a lot of time spent during revision for KCSE. It is also cost effective
MOCK Past Papers
As a student, you will have access to the most important resources that can help you understand what is required for you to sit and pass your KCSE examination and proceed to secondary school or gain entry to University admission respectively.
KCSE ONLINE
Similar
More
Similar
KCSE ONLINE WEBSITE provide KCSE, KCPE and MOCK Past Papers which play a great role in students� performance in the KCSE examination.
Choosing the KCSE mock examination revision material saves you a lot of time spent during revision for KCSE. It is also cost effective
Ask for clarification if in doubt, vitae dignissim est posuere id.
Indices and Logarithms
sit amet congue Mock Past Papers, give you an actual exam situation in readiness for your forthcoming national examination from the Kenya National Examination Council KNEC
Choosing the KCSE mock examination revision material saves you a lot of time spent during revision for KCSE. It is also cost effective sapien.
Choosing the KCSE mock examination revision material saves you a lot of time spent during revision for KCSE. It is also cost effective sapien.
Indices and Logarithms
As a supplementary to coursework content our e-library for digitized multimedia CDs while enhance and ensure that you never missed that important concept during the normal class lessons. It is a Do it Yourself Project..
Candidates who would want their papers remarked should request for the same within a month after release of the results. Those who will miss out on their results are advised to check with their respective school heads and not with the examination council
For Best results INSTALL Adobe Flash Player Version 16 to play the interactive content in your computer. Test the link below to find out if you have Adobe Flash in your computer.