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Section 1: Number Skills
1. Introduction
This section on number skills lays the foundation for the other sections, as it
provides the student with the opportunity to review basic mathematical
concepts that will be used in this subject throughout the year. As the student
you need to make sure that on completion of this section you are comfortable
working with all of its contents.
1.1
Learning outcomes
At the end of this section on Number Skills you should be able to competently
do the following:
 Fractions
Multiply and divide fractions
Add or subtract fractions with the same or different denominators
Simplify fractions
 Decimals
Convert a fraction to a decimal by using a calculator
Use the correct notation to indicate a recurring decimal
Convert a decimal to a fraction
 Rounding off
Moet nog invul
 Percentages
Convert decimals and fractions to percentages and vice versa
Do simple increasing and decreasing problem solving by using
percentages
 Ratio and Proportion
Compare information in terms of ratios
Do simple problem solving by using direct and inverse proportion
 Sigma notation
Determine sums written in sigma notation
Write sums in sigma notation
 Exponents
1
Know and be able to apply all the Laws of Exponents to simplify
expressions
 Logarithms
Moet nog invul
Start up activity 1.1: What do I know?
Pair up with a class mate and complete the following activity within 15
minutes:
Try to define each of the concepts listed below and support your definition with
an example of each. Then share your findings with the rest of the class.
 Fractions
 Decimals
 Rounding off
 Percentages
 Ratio and Proportion
 Sigma notation 
 Exponents
 Logarithms
2. Fractions
Most of us are accustomed to fractions such as “three over four”
3
, and would
4
interpret this specific fraction as “ three parts of four” or “three divided by four”.
This is indeed correct. However, fractions are also known as rational numbers
and the set of all rational numbers are indicated with the sign .
Definition of a Rational Number
A rational number is a number that can be written as a fraction
a
, where a
b
and b are integers and b  0 , as division by 0 is not defined.
We can also refer to
a
as the quotient of a and b , where a is called the
b
numerator and b the denominator or divisor of the quotient.
Example
5 20 0 4
3
,
, , and
are all examples of rational numbers.
7 9 6 1
8
2
To do different operations such as addition, multiplication and division with
rational numbers, also called fractions, we need certain properties to guide us
in working with these special numbers.
2.1
1.
Properties of Fractions
a c ac
 
b d bd
To multiply fractions, we multiply the numerators with each other and the
denominators with each other.
Example
3 7 3  7 21
 

5 2 5  2 10
2.
a c a d
  
b d b c
To divide fractions, invert the divisor and multiply it with the first fraction. This
operation becomes meaningful if we look at the following explanation:
a d ad ad

a c b c bc bc ad
 



b d c d cd
1
bc

d c dc
Example
3 7 3 2 6
   
5 2 5 7 35
Please note that we are NOT doing cross-multiplication here. Crossmultiplication is an operation which is performed ONLY if you are working with
equations, not expressions. Recall that an expression is obtained by
combining variables and constants by using operations such as addition,
subtraction, multiplication, division and roots. This is an example of an
1
3
expression: 2 x  x 2  7  4 x . An equation is a statement that two
mathematical expressions are equal. These are examples of equations:
7  4  3 and 2x  4  9 .
3.
a b ab
 
c c
c
To add fractions that have the same denominators, you only need to add the
numerators.
Example
3 6 36 9
 

5 5
5
5
3
4.
a c ad  bc
 
b d
bd
To add fractions with different denominators, find a common denominator and
rewrite the terms in the numerator with respect to the new denominator. Then
add the new terms in the numerator. This process once again becomes
meaningful if we look at the following explanation:
a c a  d c  b ad  bc
 


b d bd d b
bd
Example
3 7 3  2 7  5 6  35 41
 



5 2 5 2 25
10
10
Please note that we are once again NOT doing cross-multiplication here.
Cross-multiplication is an operation which is performed ONLY if you are
working with equations, not expressions.
5.
ac a

bc b
Factors that are common in the numerator and the denominator can be
cancelled if there is multiplication and/or division between all the factors.
Example
3 4 3

5 4 5
6.
If
a c
 , then ad  bc
b d
Now this is cross-multiplication, as we are working with an equation, not an
expression.
Example
3 9
 , because 3 15  9  5
5 15
Example
Evaluate
9 211

24 300
Solution:
When evaluating a sum consisting of fractions for which a common
denominator can be found which is smaller than the product of the two
denominators as described in property 4 above, the following procedure is
used. Each denominator is factored into prime factors (factors that can only
divide into 1 and itself). So 24  2 3  3 and 300  2 2  3  5 2 . The Least Common
4
Denominator (LCD) is then determined by forming the product of all the
factors that occur in the factorizations, using the highest power of each factor.
Therefore the LCD of 24 and 300 is 2 3  3  5 2  600 .
Then
9 211


24 300


9  25 211  2

24  25 300  2
225 422

600 600
647
600
3. Decimals
It is not always easy and appropriate to work with numbers in fractional form.
You would, for instance, be very surprised to see the price of a pair of jeans
being advertised as R129,
95
75
. Or a packet of biltong being sold for R . We
100
4
therefore often choose to express fractions as decimals.
All fractions (rational numbers) can be expressed as recurring decimals, that
is, a decimal in which one or more of the digits are repeated endlessly. A
recurring decimal is indicated by placing a dot or a line over the digit or digits
that repeat. Here are some examples of recurring decimals:
1
 0,333333  0,3
3
157
 0,317171717   0,317
495
9
 1,285714285714285714  1, 285714
7
1
 0,50000  0,50
2
The last example is also called a terminating decimal, as it can be written as a
fraction with a denominator that can be expressed as a power of 10.
1 5

 0,50000  0,50
2 10
It is customary to omit indicating the recurring decimal when working with
terminating decimals. We would, for example, rather write
1
 0,5 than
2
1
 0,50 .
2
Whenever we encounter a recurring or terminating decimal, we know that it
can be expressed as a fraction. To convert a repeating decimal such as
x  2,578787878  2,578 to rational form, we complete the following procedure:
5
Step 1:
Multiply the decimal number with an appropriate power of 10 so that the
comma moves enough spaces to past the first recurrence. In this example we
will multiply with 1000 to move the comma until after the first number 8.
1000 x  2578,7878
Step 2:
Multiply the decimal number with an appropriate power of 10 so that the
comma moves enough spaces to stop just before the first recurrence. In this
example we will multiply with 10 to move the comma until just before the first
number 7.
10 x  25,787878
Step 3:
This step involves subtracting the two equations obtained in steps 1 and 2
from each other in order to eliminate the decimals.
1000 x  2578,787878
10 x 
25,787878
990 x  2553,0
Step 4:
Divide by 990 on either side of the last equation to obtain the fractional
answer. So 2,578 
2553
.
990
4. Rounding off
Moet dit nog doen.
5. Percentages
Percentages form part of our everyday lives. They are often quoted in the
media, particularly in connection with money matters:
‘Inflation now stands at 7,5%.’
‘Ford has given its workforce a 24,2% raise.’
‘Unemployment in South Africa is about 36%.’
26% of children
watch 15 hours or
more television
per week.
6
In the sections on fractions and decimals, we learned some very useful ways
to express parts of quantities and to compare quantities. In this section we
now standardise our comparison of parts of quantities. We will do this by
expressing quantities in relation to a standard unit of 100. We use this
relationship, called percent, to solve many different types of problems.
The word percent means “per hundred” or “for every hundred”. So, if you pay
14 percent VAT on an item, it means that for every R100 you will pay R14
VAT. The symbol % represents a “percent”.
Definition of a Percent
A percent is a fractional part of a hundred expressed by a percent sign (%).
A hundred percent (100%) represents one whole quantity, or one hundred out
 100 
.
 100 
of a hundred parts, 
Using the label “percent” is an easy way to express the relationship of any
quantity to 100. When we want to make a comparison or state a problem, the
term percent may save time in performing a calculation. But we can't use
percentages as such when solving a problem. First of all, we need to convert
the percentage to a fraction or a decimal.
To change a percentage such as 45% to a numerical equivalent, we need to
divide by 100. Then, for every real number a , a% is numerically equal to
a  100 or
a
45
. So 45% 
.
100
100
Percentages often help us make comparisons between numbers.
Example
Suppose a student obtained the following marks in a series of tests:
Mathematics
English
Life Skills
16
25
31
50
30
40
It is difficult to compare the results that the student obtained in each subject,
as the tests were not out of the same mark. The student obtained the most
marks in English, but was this the best result? In order to compare the marks,
the fractions should all be converted to percentages, which will give us results
expressed as fractions with the same denominator, namely 100. We go about
doing this as follows:
7
16 4 64
 
 64%
25 4 100
31 2 62
 
 62%
50 2 100
30 2,5 75


 75%
40 2,5 100
The test results are therefore 64% in Mathematics, 62% in English and 75% in
Life Skills. The best result was obtained in Life Skills.
To convert a fraction or decimal to a percentage, it is easiest to just multiply
with 100.
Examples
Expressing a decimal as a percentage
0,59  100  59%
1.
1,45  100  145%
2.
Expressing a fraction as a percentage
3.
4.
5.
34 100

 34%
100 1
103 100

 103%
100 1
19 100 1900


 52,78%
36 1
36
Expressing a percentage as a decimal and a fraction
6.
5.1
30%  0,30 
30
100
Increasing an Amount
In real life it is often indicated that products will rise in price with a certain
percentage. We could hear on the news that bread, for example, will increase
in price with 3% as from next week. What does such a percentage increase
mean in cash terms?
Example
If bread currently costs R 7,50 and increases in price with 3%, how much will it
cost after the raise?
There are two ways in which we can go about determining the new price of
bread. Firstly, we can calculate 3% of R 7,50 and then add it to the original
price:
3% of R 7,50
3
 R 7,50
100
 23c

8
New price
 R7,50  R0,23
 R7,73
Secondly, we can consider the original amount of R 7,50 as 100% of the price.
Increasing it by 3% is the same as calculating 103% of the original cost.
New cost  103%  R7,50
103
 R 7,50
100
 R 7,73

5.2 Decreasing an Amount
Just as prices of items that we buy everyday may rise occasionally, they do
sometimes become cheaper. The amount of money that we pay for those
items then decreases.
Example
The price of a movie ticket is R 23,50 . What would the price be after the
discount as indicated on the banner is applied?
There are two ways in which we can go about
determining the new price of a movie ticket.
Firstly, we can calculate 5% of R 23,50 and
then subtract it from the original price:
5
 R 23,50
100
 R1,18
 R 23,50  R1,18
5% of R 23,50
New price
5% off on all
movie tickets

 R 22,32
Secondly, we can consider the original amount of R 23,50 as 100% of the price.
Decreasing it by 5% is the same as calculating 95% of the original price.
New cost  95%  R 23,50
95
 R 23,50
100
 R 22,32

6. Ratio and Proportion
6.1
Ratio
Definition of Ratio
A ratio compares two or more quantities of the same kind, expressed in the
same unit.
9
Example
At a BMW dealership there are 24 blue BMWs for sale and 15 silver ones. The
ratio of blue to silver BMWs can be written as 24 : 15 .
Note that the unit in this example is simply “cars” and is not included or
indicated in the ratio.
A ratio can be simplified by dividing each side of the ratio by the same
number.
Example
Cost price : Selling price  250 : 300 (No units are indicated, but have to be
the same - rands on either side)
(Divide both sides by 25)
 10 : 12
(Divide both sides again – now by 2 – to
 5:6
obtained the simplified form)
Quantities can be divided in a certain ratio.
Example
Mr Jones, Mrs Black and Dr Edwards invested in an investment opportunity
through ABSA bank and made a total profit of R1 048 238. Mr Jones invested
R180 000, Mrs Black invested R315 000 and Dr Edwards invested R135 000.
Divide the profit among the investors in the same ratio as their investments.
First we need to determine the ratio between the amounts that they invested.
Jones : Black : Edwards = 180 000 : 315 000 : 135 000 = 4 : 7 : 3
We get this simplified ratio by dividing each amount by 45 000. The sum of the
parts is 4 + 7 + 3 =14.
4
 R1 048 238 = R299 496,57
14
7
 R1 048 238 = R524 119,00
Mrs Black receives
14
3
 R1 048 238 = R224 622,43
Mr Edwards receives
14
Therefore: Mr Jones receives
6.2 Proportion
Definition of Direct Proportion
Two quantities are in direct proportion if the one quantity increases
(decreases) as the other quantity increases (decreases).
10
These two quantities, which we can represent by x and y , are in a functional
relationship that can be indicated by:
1.
an equation in the form y  mx such as y  4 x . Then:
y

x
-2 -8
-1 -4 Here we can clearly see how the y  values increase as the
0
0
x  values increase.
1
4
2
8
3
12
11
2.
a straight line graph going through the origin (0, 0).
y
12
2
2
28
2
2
2
4
2
2
21
2
2
2
12
2
-2
-1
1
2
3
x
-4
-8
This straight line graph is
representative of the equation
y  4x .
Example
4 kg of potatoes cost R12,50 . How much does 11 kg cost?
If 4 kg cost R12,50 , then
4
12,50
 R3,13 .
 1 kg costs
4
4
Therefore 11 kg of potatoes will cost 11  R3,13  R34,43
Definition of Inverse Proportion
Two quantities are in inverse proportion if the one quantity increases as the
other quantity decreases.
These two quantities, which we can represent by x and y , are also in a
functional relationship to each other and can be indicated by:
12
k
x
1.
an equation in the form xy  k or y 
If y 
3
, then we have that
x
y
3

4
3
  1
3
3
Here we can clearly see how the y  values decrease

2
3
3
  3
from  to  3 as the x  values increase from  4 to  1 .
1
4
3
Division by 0 is not defined.
 Undefined
0
3
3
1
3
3
It is also clear that the y  values decrease from 3 to
2
4
3
as the x  values increase from 1 to 4.
1
3
3
4
x
-4
-3
-2
-1
0
1
2
3
4
or
13
2.
a graph of a hyperbola with center at (0, 0).
y
3
2
2
22
4
2
21
2
2
2
2
12
2
-4
-3
-2
-1
1
2
3
4
x
-1
11
4
-2
-3
This hyperbola is representative of the equation y 
3
.
x
Example
9 workers are asked to paint the inside walls of an apartment building and
they are given 2 weeks to do it in. How many days will it take to paint the
building if 15 workers are doing the job?
This is an inverse proportion problem as the number of days that it would take
to paint the building will surely be less if more painters are employed to do the
job. We will therefore use the equation xy  k and first determine k .
The values of x and y are:
x  9 workers and y  14 days (two weeks)
 k  xy  9  14  126
If the number of workers (x ) increases to 15, we then have that
y
k
x
126

15
 8,4 days to paint the inside walls of the apartment building.

14
7. Sigma Notation
Summation or sigma notation is basically shorthand for writing sums. The
sums that can be written in this way are special. The terms that make up such
a sum have to belong to a generated sequence and are written in a specific
order. The sign used in sigma notation is derived from the Greek letter  ,
which corresponds to our S for “sum”.
If we want to add the terms a1 , a2 , a3 ,, an , sigma notation can be used to
n
express the sum as
a
k 1
k
 a1  a 2  a3    a n .
The letter k is called the index of summation and a k is the general term of
the sum. The idea is to substitute the values of k into the general term to
obtain the sum on the right-hand side of the equation. The index of summation
indicates that the first value we should substitute is k  1 and the last value is
k  n.
Example
To find the first three terms, the 20th term and the 598th term of the sequence
defined by the general term ak  2k  15 , we have to substitute 1 , 2 , 3 , 20 and
598 into the formula representing the general term. Then we get that
the first term a1  2(1)  15  13 ,
the second term a2  2(2)  15  11 ,
the third term a3  2(3)  15  9 ,
the twentieth term a20  2(20)  15  25 and
the five hundred and ninety eight term a598  2(598)  15  1181 .
If the sum of a number of terms is expressed in sigma notation, we can
determine the sum as demonstrated in the following three examples.
Example
7
Determine the sum
k
2
.
k 1
7
k
2
 12  2 2  3 2  4 2  5 2  6 2  7 2
k 1
 1  4  9  16  25  36  49
 140
Example
5
Determine the sum
3
k
k 1
2
.
15
5
3
k
k 1
2
3
3
3
3
3
 2  2  2  2
2
1
2
3
4
5
3 1 3
3
 3   
4 3 16 25
3600 900
400
225 144





1200 1200 1200 1200 1200
5269

1200

Example
6
Determine the sum
5 .
k 1
6
5  5  5  5  5  5  5
k 1
 5 6
 30
This summation means that the value 5 must be
added 6 times.
To determine how many terms are in a given summation, subtract the bottom
6
index from the top index and add 1 . In the summation
5
the number of
k 1
terms will therefore be N  6  1  1  6 terms. In the summation
12
 4t
the
t 3
number of terms will be N  12  3  1  10 .
As suggested in the previous sentence, the index of a summation does not
have to start at k  1 . It is quite possible to determine the sum of
12
 4t , as
t 3
illustrated in the next example.
Example
12
Determine the sum
 4t .
t 3
12
 4t
t 3
 4(3)  4(4)  4(5)  4(6)  4(7)  4(8)  4(9)  4(10)  4(11)  4(12)
 12  16  20  24  28  32  36  40  44  48
 300
Just as you should be able to determine the value of a summation written in
sigma notation, you should be able to express a sequence of terms using
sigma notation.
16
Example
Express 12  2 2  32  4 2 in sigma notation. Then
4
k
2
 12  2 2  3 2  4 2 .
k 1
Notice that it is not compulsory to use the letter k to represent the index. Any
alphabet letter can be used. A certain preference is given to the letters k and
n , however. The previous example can thus also be written as
4
n
2
 12  2 2  3 2  4 2 , without changing the meaning of the summation or the
n 1
value of the sum.
Example
Write the sum
5  7  9    21 using sigma notation.
First we need to determine the general term a k , that represents all of the
terms in the sum. It seems as if it is ak  2k  1 , because all the values under
the root signs are odd values. Next we need to determine the starting and
ending points of the index of summation, k . The starting point has to be k  3 ,
as 2k  1  2(3)  1  5 . The ending point will be k  11 , because
2k  1  2(11)  1  21 .
11
Then
5  7  9    21   2k  1 .
k 3
But there is no unique way to write a sum in sigma notation. The answer we
obtained in the previous example is just one way of writing the sum
5  7  9    21 in sigma notation. It all depends on the choice of index,
which in turn will influence how the general term is expressed.
If we choose to let k start at 4 instead of 3 , we obtain the following notation:
12
5  7  9    21   2k  3
k 4
The index of summation now runs from 4 to 12 and the general term is
expressed by ak  2k  3 . The terms obtained by these indices and general
11
term are, however, exactly the same as those represented by

2k  1 .
k 3
8. Exponents
In this section we recall information regarding integer exponents a m  , radicals
x and n th roots n a in order to give meaning to an exponent in which the
 
 
m
power is a fraction  a n  .


17
8.1
Integer Exponents
Instead of writing out a product of identical numbers as 3  3  3  3  3 , it can be
written in exponential form, as 35 .
Definition of an Exponent
If a is any real number and n is a positive integer, then the n th power of a is
a n  a 
a 
a

a 

 a
a .

n times
The number a is called the base and n the exponent.
Power
an
Exponent
Base
In order to work with exponents, we need to know the laws according to which
they operate.
8.1.1 Laws of Exponents
1.
a m a n  a mn
To multiply powers with the same bases, keep the base and add the
exponents; then a 4 a 6  a 46  a10 .
Proof: If a is any real number and m and n positive integers, then
a m a n  (a
a

 a)( a
a

 a)  a 
a 
a

a
 a  a m n


 
m factors n factors
m  n factors
2.
am
 a mn
n
a
To divide powers with the same bases, keep the base and subtract the
exponent in the denominator from the exponent in the numerator; then
a6
 a 6 4  a 2 .
4
a
Proof: If a is any real number and m and n positive integers, then
factors
m


a m a  a  a  a  a

 a
a
 
 a  a mn


n
a 
a
 a 

a
a
m  n factors
n factors
3.
a 
m n
 a mn
18
To raise a power to a new power, keep the base and multiply the
3
exponents with each other; then a 6   a 63  a18 .
4.
ab n
 a nb n
To raise a product to a power, raise each factor of the product to the
power; then ab 4  a 4 b 4 .
n
5.
an
a
   n
b
b
To raise a quotient to a power, raise both the numerator and the
6
a6
a
denominator to the power; then    6 .
b
b
6.
a
 
b
n
b
 
a
n
To raise a fraction to a negative power, invert the fraction and change
a
the sign of the exponent; then  
b
7.
2
2
b
  .
a
a n b m

b m a n
When moving a power from the numerator to the denominator or vice
versa, change the sign of the exponent; then
a n b m
.

b m a n
There are two more important definitions that follow from the Laws of
Exponents, namely:
8.
If a  0 is any real number and n is a positive integer, then a 0  1 .
Proof: a 0  a 4  a 0 4  a 4 . But this is only possible if a 0  1 .
9.
If a  0 is any real number and n is a positive integer, then a  n 
Proof: a n  a  n  a n  (  n )  a 0  1 . But this is only possible if a  n 
1
.
an
1
.
an
Examples
Eliminate negative exponents and simplify as far as possible:
x 4 x 7  x 4 7  x11
1.
19
2.
x 3 x 9  x 3 ( 9 )  x 6 
3.
y 12
 y 125  y 7
5
y
4.
5.
c 
2 7
5 x 2
1
x6
 c 27  c14
 5 2  x 2  25 x 2
3
6.
7.
8.
h3 h3
h
   3 
27
3
3
x 5
1
 2 5
2
y
y x
4a c  7ac   4 a  c  7ac 
 16a c 7ac 
9 3 2
6
9 2
2
3 2
18 6
6
6
 (16)(7)a18 ac 6 c 6
 112a 19 c12
9.
4 s 2 t 3 k 6
12t 4 sk 
4
1s 2 sk 
3t 4 t 3 k 6
4

s2s4k 4
3t 7 k 6
s6
 7 2
3t k

10.
x
 
 y
3

 y 2 x3

 z 2


4
5
  y  3  y 8 x12  5
   

  x   z  2 

3


5
 y
   y 8 x12 z 2
x
 y3 
  3  y 40 x 60 z 10
x 
 y 3 y 40 x 3 x 60 z 10


 y 43 x 57 z 10
8.2
Radicals (Roots)
Ek moet dit nog doen.
20
8.3
Rational Exponents
Ek moet dit nog doen.
9. Logarithms
The study of logarithms is motivated by the need to solve equations such as
bx  a
for x. To be able to express x in terms of a and b in this setting requires a
manipulation not yet presented. In fact, a new notation must be introduced in
order to solve this problem. In view of this, we define x  log b a to mean the
same as b x  a .
Logarithm Notation
x  log b a means the same as b x  a where b  0, b  1, and x is any real
number.
The equation x  log b a is read as “x is the logarithm of a to the base b” or as
“x is the logarithm to the base b of a.” Notice that the logarithm is equal to x,
and x is the exponent. Thus, a logarithm is an exponent. In this case, x is the
exponent to which b must be raised to produce a.
The number a results from raising the positive base b to some power x.
Consequently, a is always positive in b x  a . In turn, this means that a must be
positive in x  log b a . In other words, log b a is defined only for a  0 .
The logarithm of a negative number (or zero) is not defined.
Here are some examples of equations written in logarithmic form along with
their equivalent exponential form.
Logarithmic equation
Exponential equation
log 3 9  2
log 2 8  3
log 4 1  0
log 10 0,01  2
log e 1  0
log 10 ( 2)  undefined
32  9
23  8
40  1
102  0,01
e0  1
10 ?  2
? there is no possible
answer
21
The two most popular bases for logarithms are 10 and e. Logarithms using
base 10 are called common logarithms, and “ log 10 ” is written simply “log”.
When no base is written, base 10 is assumed. log 10 x is written log x
Example
Solve the following:
(a) log 2 64  y
1
y
243
1
(c) log 49 x  
2
(b) log 3
Solutions
(a)
2 y  64
y6
(b)
1
 3 5
5
3
y  5
3y 
(c)
x  49
x

1
2
1
1

49 7
Assessment Activity 1.1
Solve the following:
1. y  log 9 27
2. log b 8 
3
4
2
3
4. y  log 27 3
1
 7
5. log b
128
1
6. log 1 x 
4
16
3. log 8 x  
22
9.1
The Laws of Logarithms
From our knowledge of logarithms we have
log 2 8  log 2 16  log 2 2 3  log 2 2 4  3  4  7  log 2 128  log 2 8(16)
Thus
log 2 8(16)  log 2 8  log 2 16 .
It turns out that this equation is a special case of the first law of logarithms.
Laws of Logarithms
If M and N are positive, b  0, and b  1, then
(i) log b MN  log b M  log b N
M
 log b M  log b N
N
(iii) log b ( N k )  k log b N
(ii) log b
(iv) log b (b x )  x
(v) b log x  x
In addition, log 1  0 because 10 0  1
OR: log b 1  0 because b 0  1
b
Example
Use the properties of logarithms to expand each expression.
x2
x 5
(b) log 32 x
(a) log
(c) log[ e 3 ( x)]
(d) log b
3
2x
Solution
(a) Use property (ii)
log
x2
 log( x  2)  log( x  5)
x 5
(b) Use property (iii)
log 32 x  2 x log 3
(c) Use property (i)
log[ e 3 ( x)]  log e 3  log x  3 log e  log x
(d) Use property (i) and (ii)
log b 3  log b 2 x
23
 log b 3  log b 2  log b x 
 log b 3  log b 2  log b x
Example
Use the properties of logarithms to simplify each of the following expressions:
(a) log 3  log x
(b) log x  log 2
1
2
(c) 2 log b ( x  1)  log b x
Solution
(a) Use property (I)
log[ 3( x)]
(b) Use property (ii)
log
x
2
(c) Use property (iii)
log b ( x  1)  log b x
2
1
2
Use property (i)
 12 
 log b [( x  1)  x ]
 
2
Assessment Activity 1.1
Use the properties of logarithms to simplify each of the following expressions:
1. log b x  log b x  log b 3
2. log x  log( x  1)  2 log( x  2)
3. log 4  log 10  2
4.
log A 2  log A
1
log B  log B
2
5. 2 log A  3 log B 
log A
2
Learning activity 2.1 [or Portfolio Activity 2.1]
24
Dui vir my aan waar jy dink ek ‘n Learning Activity (oefening) kan invoeg.
Group Activity 1.1
Dui vir my aan waar jy dink ek ‘n Group Activity (oefining) kan invoeg.
10. End of section comments
This wraps up all the content that you had to acquire before being able to
proceed to the next sections. Make sure that you understand and can
confidently work with everything that was addressed in this section. We will
now proceed to the next section which is on calculator usage.
11. Feedback
Hierdie sal ek invoeg as ek die Learning en Group Activities klaar geskryf het.
12. Tracking my progress
You have reached the end of this section. Check whether you have achieved
the learning outcomes for this section.
Learning outcomes
 I feel
confident
 I still need
more practice
Multiply and divide fractions
Add or subtract fractions with the same
or different denominators
Convert a decimal fraction to a fraction
with a numerator and denominator
Convert decimals and fractions to
percentages and vice versa
Do simple increasing and decreasing
problem solving by using percentages
Compare information in terms of ratios
Do simple problem solving by using
25
direct and inverse proportion
Write sums in sigma notation
Determine sums written in sigma
notation
Know and be able to apply all the Laws
of Exponents to simplify expressions
What did you like best about this section?
What did you find most difficult in this section?
What do you need to improve on?
How will you do this?
26