The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.

## Chapter1

Related Publications

Discover the best professional documents and content resources in AnyFlip Document Base.

# Chapter1

### Chapter1

1CHAPTER Indices

What will you learn?

1.1 Index Notation

1.2 Law of Indices

WWhhyy ddoo yyoouu lleeaarrnn tthhiiss cchhaapptteerr??
• WWhryitidngoaynouumlbeearrinn tinhdiesxcnhoataptitoenr?enables the

number stated in a simple and easily understood
form. Various operations of mathematics that
involve numbers in index notation can be
performed by using laws of indices.
• Concept of index is used in the fields of science,
engineering, accounting, finance, astronomy,
computer and so on.

Kenyir Lake, located in the district of Hulu
Terengganu, in Terengganu, is the biggest
man-made lake in Southeast Asia. Kenyir Lake is a
world famous tourist destination known for its unique
natural beauty. Kenyir Lake is an important water
catchment area. Kenyir Lake, which was built in
the year 1985, supplies water to Sultan Mahmud
Power Station. The estimated catchment area at the
main dam is 2 600 km2 with a reservoir volume of
13 600 million cubic metre. During rainy season,
the volume of water in the catchment area will increase
sharply. What action should be taken to address this
situation?

EExxpplloorriinngg EErraa

Index notation is an important element in the
development of mathematics and computer
programming. The use of positive indices was
introduced by Rene Descartes (1637), a well-known
French mathematician. Sir Isaac Newton, another
well-known British mathematician, developed the
field of index notation and introduced negative
indices and fractional indices.

http://bukutekskssm.my/Mathematics/F3/
ExploringEraChapter1.pdf

WORD B A N K

• base • asas

• factor • faktor

• index • indeks

• fractional index • indeks pecahan

• power • kuasa

• root • punca kuasa

• index notation • tatatanda indeks

Saiz sebenar

1

1.1 Index Notation

CHAPTER 1 What is repeated multiplication in index form? LEARNING
STANDARD
The development of technology not only makes most of our daily tasks
easier, it also saves cost of expenses in various fields. For instance, Represent repeated
the use of memory cards in digital camera enables users to store multiplication in index form
photographs in a large number and to delete or edit unsuitable and describe its meaning.
photographs before printing.
DISCUSSION CORNER

Discuss the value of the
capacity of a pen drive.

BULLETIN

The nuclear fission of
uranium U-320 follows the
pattern 30, 31, 32, …

In the early stage, memory cards were made with a capacity of 4MB. The capacity was
increased with time and the needs of users. Did you know that the value of capacity of memory
cards is calculated using a special form that is 2n?

In Form 1, you have learnt that 43 = 4 × 4 × 4. The number 43 is written in index notation, 4
is the base and 3 is the index or exponent. The number is read as ‘4 to the power of 3’.

Hence, a number in index notation or in index form can be written as;

an Index
Base

You have also learnt that 42 = 4 × 4 and 43 = 4 × 4 × 4. For example;

4 × 4 = 4 2 The value of index is 2

Repeated two times The value of index is the same as the number of times
4 is multiplied repeatedly.

4 × 4 × 4 = 4 3 The value of index is 3

R epeated three times The value of index is the same as the number of times
4 is multiplied repeatedly.

Example 1

Write the following repeated multiplications in index form an. REMINDER

(a) 5 × 5 × 5 × 5 × 5 × 5 (b) 0.3 × 0.3 × 0.3 × 0.3 25 ≠ 2 × 5 43 ≠ 4 × 3
an ≠ a × n
Saiz se ((bec))e m(n–a2×r) m × (×–2m) ×× m(– 2×) m × m × m ((fd)) n—41× ×n—41× n××—41n ×× —n41 ××n—41× n × n

2

Chapter 1 Indices

Solution:

(a) 5 × 5 × 5 × 5 × 5 × 5 = 56 (b) 0.3 × 0.3 × 0.3 × 0.3 = (0.3)4 CHAPTER 1

repeated six times repeated four times

( )( c) (–2) × ( –2) × (–2) = (–2)3 (d) —14 × —14 × —41 × —14 × —14 = —41 5
repeated three times
repeated five times

(e) m × m × m × m × m × m × m = m7 (f) n × n × n × n × n × n × n × n = n8

repeated seven times repeated eight times

From the solution in Example 1, it is found that the value of index in an index form is the same as
the number of times the base is multiplied repeatedly. In general,

an = a × a × a × … × a ; a ≠ 0
n factors

MIND TEST 1.1a
1. Complete the following table with base or index for the given numbers or algebraic terms.

53 (– 4)7 Base Index
5 7
( ) —21 10 ( )m6 – —37 4 —21
n 6
n0 (0.2)9 9
x 4
( )x20 2 —31 2 8 2

8

2. State the following repeated multiplications in index form an.

(a) 6 × 6 × 6 × 6 × 6 × 6 (b) 0.5 × 0.5 × 0.5 × 0.5 × 0.5 × 0.5 × 0.5

(c ) —21 × —21 × —21 × —12 (d) (–m) × (–m) × (–m) × (–m) × (–m)

( ) ( ) ( ) ( ) ( ) ( ) ( e ) 1 —23 × 1—23 × 1—23 ( f) – 1n– × – 1n– × – –1n × – 1n– × – –1n × – 1n–

3. Convert the numbers or algebraic terms in index form into repeated multiplications.

( ) ( ) (a) ( –3)3 (b) (2 .5 )4 (c) —23 5 ( d) – 2 —14 3

( ) (e) k 6 (f) (– p) 7 (g) —m1 8 (h) (3n)5 Saiz sebenar

3

CHAPTER 1 How do you convert a number into a number in index LEARNING
form? STANDARD

A number can be written in index form if a suitable base is selected. You Rewrite a number in index
can use repeated division method or repeated multiplication method to form and vice versa.
convert a number into a number in index form.

Example 2 FLASHBACK
Write 64 in index form using base of 2, base of 4 and base of 8.
Solution: 4 × 4 × 4 = 43

Repeated Division Method

(a) Base of 2 (b) Base of 4 (c) Base of 8
• 64 is divided repeatedly • 64 is divided • 64 is divided
by 2. repeatedly by 4. repeatedly by 8.

2 ) 64 n=3 4 ) 64 n=2 8 ) 64
2 ) 32 4 ) 16 8 ) 8
2 ) 16 4 ) 4
n=6 2 ) 8 1 1
2 ) 4
2 ) 2 Hence, 64 = 82
1
Hence, 64 = 43

The division is
continued until
1 is obtained.

Hence, 64 = 26

Repeated Multiplication Method

(a) Base of 2 (b) Base of 4 (c) Base of 8
2×2×2×2×2×2 4×4×4 8 × 8 = 64
4 16
8 64 Hence, 64 = 82
16
32 Hence, 64 = 43 DISCUSSION CORNER
64
Which of the repeated
Hence, 64 = 26 division method and the
repeated multiplication
Saiz sebenar method is easier to
convert a number into a
4 number in index form?
Discuss.

Example 3 Chapter 1 Indices CHAPTER 1

W rite —3 31—225 – in index form using base of —52 . Repeated Multiplication Method
Solution: — 52 × —25 × —52 × —52 × —25
—245–
Repeated Division Method 1—285–
6—1265–
n=5 2 ) 32 n=5 5 ) 3 125 — 331—225–
2 ) 16 5 ) 625
2 ) 8 5 ) 125 ( )Hence, —331—22 5– = —52 5
2 ) 4 5 ) 25
2 ) 2 5 ) 5
1 1

( )Hence, —331—22 5– = —52 5

MIND TEST 1.1b

1. Write each of the following numbers in index form using the stated base in brackets.

( a) 8 1 [base of 3] ( b) 15 6 25 [base o f 5] (c) —1624–5 [ ]base of —45

[ ( )] (d) 0 .00032 [ base of 0 .2] (e) – 16 38 4 [base of (– 4)] ( f) —116
base of – —14

How do you determine the value of the number in index form , an?

The value of an can be determined by repeated multiplication method or using a scientific
calculator.

Example 4 QU I Z
Calculate the values of the given numbers in index form.

(a) 25 (b) (0.6)3 (m)4 = 16
What are the possible
2 × 2 × 2 × 2 × 2 0.6 × 0.6 × 0.6 values of m?

4 × 2 0.36 × 0.6

8 × 2 0.216
16 × 2 0.63 = 0.216

32 Saiz sebenar
Hence, 25 = 32 Hence, 0.63 = 0.216

5

Example 5 SMART FINGER 1,234567.89 REMINDER

78 9÷ Negative or fractional base
45 6x must be placed within
12 3- brackets when using a
AC 0 .+ calculator to calculate
values of given numbers.
CHAPTER 1 (a) 54 = 625 5 ^ 4 =
DISCUSSION CORNER
(b) (–7)3 = –343 ( (–) 7 ) ^ 3 =
Calculate questions (c),
( )( c) —32 4 —8116– ( 2 ab/c 3 ) ^ 4 = (d) and (e) in Example 5
without using brackets.
same? Discuss.
( ) ( d) 1—35 2= —6245– ( 1 ab/c 3 ab/c 5 ) ^ 2 =
( (–) 0 . 5 ) ^ 6 =
(e) (– 0.5)6 = 0.015625

MIND TEST 1.1c

1. Calculate the value of each of the following numbers in index form.

( ) ( ) ( ) ( ) (( ae )) 9 —438 5 (( bf)) ( –– 4—16) 5 4 (( cg )) (21. 5—23) 3 2 ((hd )) (–– 32. 2—13) 3 3

1.2 Law of Indices

What is the relationship between multiplication of LEARNING
numbers in index form with the same base and STANDARD

repeated multiplication? Relate the multiplication
of numbers in index
Brainstorming 1 In pairs form with the same
base, to repeated
Aim: To identify the relationship between multiplication of multiplications, and hence
make generalisation.
numbers in index form with the same base and repeated

multiplication.

Steps:

1. Study example (a) and complete examples (b) and (c).

2. Discuss with your friend and state three other examples.

3. Exhibit three examples in the mathematics corner for other groups to give feedback.

Multiplication of Repeated multiplication
numbers in index form

(a) 23 × 24 3 factors 4 factors 7 factors (overall)

(2 × 2 × 2) × (2 × 2 × 2 × 2) = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27

23 × 24 = 2 7 7=3+4
23 × 24 = 2 3 + 4

(b) 32 × 33 2 factors 3 factors 5 factors (overall)

Saiz sebenar (3 × 3) × (3 × 3 × 3) = 3 × 3 × 3 × 3 × 3 = 35

32 × 33 = 3

32 × 33 = 3

6

Chapter 1 Indices

Multiplication of Repeated multiplication
numbers in index form

(c) 54 × 52 4 factors 2 factors 6 factors (overall) CHAPTER 1

(5 × 5 × 5 × 5) × (5 × 5) = 5 × 5 × 5 × 5 × 5 × 5 = 56
54 × 52 = 5
54 × 52 = 5

Discussion:

What is your conclusion regarding the relationship between multiplication of numbers in index
form and repeated multiplication?

From Brainstorming 1, it is found that;

23 × 24 = 23 + 4 DISCUSSION CORNER
32 × 33 = 32 + 3
54 × 52 = 54 + 2 Given,
am × an = bm × bn.

Is a = b? Discuss.

In general, am × an = a m + n
Example 6

Simplify each of the following.

( a) 72 × 7 3 (b) (0 .2 )2 × (0. 2 )4 × (0. 2) 5 (c ) 2k2 × 4 k3 (d) 3m4 × —16 m5 × 12m
Solution:

(a) 72 × 73 (b) (0.2)2 × (0.2)4 × (0.2)5 REMINDER

= 72 + 3 = (0.2)2 + 4 + 5

= 75 = (0.2)11 a = a1

( c) 2=k(22× × 44k)3( k2 × Okp3e)ratio n o f ( d) 3 = m( 43 ××—16—16m×5 × 12m × m5 × m1)
12) (m4

= 8k2 + 3 the coefficients = 6m4 + 5 + 1 SMART MIND

= 8k5 = 6m10 If ma × mb = m8, such

that a > 0 and b > 0,

MIND TEST 1.2a what are the possible

values of a and b?

1. Simplify each of the following.

(a) 32 × 3 × 34 (b) (– 0.4)4 × (– 0.4)3 × (– 0.4)

( ) ( ) ( ) ( ) ( ) ( ) ( c) —74 × —74 3 × —74 5 (d) – 1—52 2 × – 1—52 3 × – 1—52 5

(e ) 4m2 × —12 m3 × (– 3)m 4 (f) n6 × —245 n2 × —54 n3 × n Saiz sebenar
( g ) – x4 × —245 x × 1—52 x2 (h) – —21 y5 × (– 6)y3 × —31 y4

7

CHAPTER 1 How do you simplify a number or an algebraic term in index TIPS
form with different bases?
Group the numbers or
Example 7 algebraic terms with the
Simplify each of the following. same base first. Then add
the indices for the terms
with the same base.

(a) m3 × n2 × m4 × n5 (b) (0.3)2 × (0.2)2 × 0.3 × (0.2)5 × (0.3)3

( c) p2 × m3 × p4 × n3 × m4 × n2 (d) –m4 × 2n5 × 3m × —14 n2

Solution:

(a) m3 × n2 × m4 × n5 (b) (0.3)2 × (0.2)2 × 0.3 × (0.2)5 × (0.3)3

= m3 × m4 × n2 × n5 wGirtohutphtehse atmerembsase. = (0.3)2 × (0.3)1 × (0.3)3 × (0.2)2 × (0.2)5
= m3 + 4 × n2 + 5 = (0.3)(2 + 1 + 3) × (0.2)(2 + 5)

= m7 × n7 Add the ind ices for terms = (0.3)6 × (0.2)7
with the same base.
= m7n7

(c) p2 × m3 × p4 × n3 × m4 × n2 (d) –m4 × 2n5 × 3m × —41 n2

= m3 × m4 × n3 × n2 × p2 × p4 = ( –1 × 2 × 3 × —41 ) m4 × m1 × n5 × n2
= m3 + 4 × n3 + 2 × p2 + 4 = – —23 m4 + 1 n5 + 2 REMINDER
= m7 n5 p6

= – —23 m5 n7 –an ≠ (–a)n
Example:
MIND TEST 1.2b –32 ≠ (–3)2
–9 ≠ 9

1. State in simplest index form.

(a) 54 × 93 × 5 × 92 (b) (0.4)2 × (1.2)3 × (0.4) × (1.2)5 × (1.2)

( c ) 1 2x5 × y3 × —12 x × —23 y4 (d) –2k5 × p6 × —14 p5 × 3k

What is the relationship between division of numbers in LEARNING
index form with the same base and repeated multiplication? STANDARD

Brainstorming 2 In pairs Relate the division of
numbers in index form
Aim: To identify the relationship between division of numbers in with the same base, to
index form with the same base and repeated multiplication. repeated multiplications,
and hence make
generalisation.

Steps:

1. Study example (a) and complete examples (b) and (c).

Saiz seb32e.. naPDrriessceunsts with your friend and state three other examples.

8

Chapter 1 Indices

Division of numbers Repeated multiplication
in index form
5 factors CHAPTER 1
(a) 45 ÷ 42
— 4452 = — 4 ×—4—×4—×4 ×—4 4––×–4– = 4 × 4 × 4 = 43
(b) 26 ÷ 22
3 factors (Remainder)
(c) (–3)5 ÷ (–3)3
2 factors

45 ÷ 42 = 4 3 3=5–2

45 ÷ 42 = 4 5–2

6 factors

—2262 = — 2 ×—2—×—22 ×—× 22––×–2–—× 2– = 2 × 2 × 2 × 2 = 24

4 factors (Remainder)

2 factors

26 ÷ 22 = 2
26 ÷ 22 = 2

5 factors

— (( ––33—))35 = ( —–3—) ×—((–—–33)—) ××–((––––33–))—××(–(–––33–))— ×—(––3–) = (–3) × (–3) = (–3)2

2 factors (Remainder)

3 factors

(–3)5 ÷ (–3)3 = (–3)

(–3)5 ÷ (–3)3 = (–3)

Discussion:

What is the relationship between division of numbers in index form and repeated
multiplication?

From Brainstorming 2, it is found that; SMART MIND

45 ÷ 42 = 45 – 2 Given ma – b = m7 and
26 ÷ 22 = 26 – 2 0 < a < 10. If a > b,
(–3)5 ÷ (–3)3 = (–3)5 – 3 state the possible values
of a and b.

In general, am ÷ an = am – n

Example 8

Simplify each of the following.

(a) 54 ÷ 52 (b) (–3)4 ÷ (–3)2 ÷ (–3) (c) m4n3 ÷ m2n
(d) 25x2y3 ÷ 5xy (e) 12m10 ÷ 4m5 ÷ m2 (f) –16p8 ÷ 2p5 ÷ 4p2
Solution:

(a) 54 ÷ 52 (b) (–3)4 ÷ (–3)2 ÷ (–3) (c) m4n3 ÷ m2n

= 54 – 2 = (–3)4 ÷ (–3)2 ÷ (–3)1 = m4n3 ÷ m2n1

= 52 = (–3)4 – 2 – 1 = m4 – 2 n3 – 1

= (–3)1 = m2 n2 Saiz sebenar

= –3

9

(d) 25x2y3 ÷ 5xy (e) 12m10 ÷ 4m5 ÷ m2 (f) –16p8 ÷ 2p5 ÷ 4p2

CHAPTER 1 = 25x2y 3 ÷ 5 x 1y1 = —142 (m10 ÷ m 5 ÷ m 2) = –—216– (p8 ÷ p5) ÷ 4p2
= == —5525xx51yxy2 22 – 1cOoype 3e f rf–iact1iieonn tso f the === 333m(mm 53 1 –0–25 ) ÷ m 2 === –– –88 —pp8438–(÷5p÷43 p÷42pp22)

= –2p3 – 2

= –2p1

= –2p

MIND TEST 1.2c

1. Simplify each of the following. (c) —mm84—nn6
(f) –25h4 ÷ 5h2 ÷ h
(a) 45 ÷ 44 (b) 710 ÷ 76 ÷ 72

(d) —297x—x34y–y2–5 (e ) m 7 ÷ m2 ÷ m4
2. Copy and complete each of the following equations.

(a) 8 ÷ 84 ÷ 83 = 8 (b) m4n ÷ m n5 = m2n

( c) — m1—0 n—m4 ×7—nm —— n2= m 5n ( d) —27—x3y—6x×2—yx3—y – = 3x y5

3. If —224x—×× —33y2 = 6, determine the value of x + y.

What is the relationship between a number in index form LEARNING
raised to a power and repeated multiplication? STANDARD

Brainstorming 3 In pairs Relate the numbers in
index form raised to a
Aim: To identify the relationship between a number in index form power, to repeated
raised to a power and repeated multiplication. multiplication, and hence
make generalisation.

Steps:

1. Study example (a) and complete examples (b) and (c).

2. Discuss with your friend and state three other examples.

Index form raised Repeated multiplication in index form Conclusion
to a power
4 factors (32)4 = 32(4)
(a) (32)4 = 38
32 × 32 × 32 × 32
Saiz sebenar
= 32 + 2 + 2 + 2 2 is added 4 times

4 times

= 32(4)

10

Chapter 1 Indices

Index form raised Repeated multiplication in index form Conclusion
to a power
3 factors (54)3 = 5 CHAPTER 1
(b) (54)3 = 5
54 × 54 × 54
= 54 + 4 + 4 4 is added 3 times (43)6 = 4
= 4
3 times

= 54(3)

(c) (43)6 6 factors

43 × 43 × 43 × 43 × 43 × 43

= 43 + 3 + 3 + 3 + 3 + 3 3 is added 6 times

6 times

= 43(6)

Discussion:

What is your conclusion regarding the index form raised to a power and repeated multiplication
in index form?

The conclusion in Brainstorming 3 can be checked using the following method.

Example (a) Example (b) Example (c)

(32)4 = 32 × 32 × 32 × 32 (54)3 = 54 × 54 × 54 (43)6 = 43 × 43 × 43 × 43 × 43 × 43
= 32 + 2 + 2 + 2 = 54 + 4 + 4 = 43 + 3 + 3 + 3 + 3 + 3
= 38 = 512 = 418

32(4) = 32 × 4 54(3) = 54 × 3 43(6) = 43 × 6
= 38 = 512 = 418

From Brainstorming 3, it can be found that; SMART MIND

Given,
(32)4 = 32(4) mrt = 312
(54)3 = 54(3)
(43)6 = 43(6) What are the possible
values of m, r and t
In general, (am)n = amn if r > t ?

Example 9

1. Simplify each of the following.

(a) (34)2 (b) (h3)10 (c) ((–y)6)3

2. Determine whether the following equations are true or false. Saiz sebenar
(a) (42)3 = (43)2 (b) (23)4 = (22)6 (c) (32)6 = (272)4

11

Solution:

CHAPTER 1 1. (a) (34)2 (b) (h3)10 (c) ((–y)6)3
= (–y)6(3)
= 34(2) = h3(10) = (–y)18
= 38 = h30

2. (a) (42)3 = (43)2 (b) (23)4 = (22)6 (c) (32)6 = (272)4

left right left right left right

Left: Left: Left:

(42)3 = 42(3) = 46 (23)4 = 23(4) = 212 (32)6 = 32(6) = 312

Right: Same Right: Same Right:

(43)2 = 43(2) = 46 (22)6 = 22(6) = 212 (272)4 = (33(2))4 Not the

same
Hence, (42)3= (43)2 Hence, (23)4 = (22)6 = 36(4)

is true. is true. = 324

Hence, (32)6 = (272)4

is false.

MIND TEST 1.2d

1. Use law of indices to simplify each of the following statements.

(a) (125)2 (b) (310)2 (c) (72)3 (d) ((– 4)3)7
(h) ((–c)7)3
(e) (k8)3 (f) (g2)13 (g) ((–m)4)3
(d) – (72)4 = (– 492)3
2. Determine whether the following equations are true or false.

(a) (24)5 = (22)10 (b) (33)7 = (272)4 (c) (52)5 = (1252)3

How do you use law of indices to perform operations of multiplication and division?

(am × bn)q
= (am)q × (bn)q (ambn)q = amq bnq
= amq × bnq

(am ÷ bn)q ( ) —abmn– q = —abmnq–q
= (am)q ÷ (bn)q
= amq ÷ bnq

Example 10

1. Simplify each of the following.

(a) (73 × 54)3 (b) (24 × 53 × 112)5 (c) (p2q3r)4 (d) (5m4n3)2

( ) ( )Saiz se be n(ea) r — 3225 4 (f ) —23xy–37 4 ( g) —(36m—m2n—3n3) –3 (h ) —(2x—3y3—46)4x—1×0y—(132—xy2—)3

12

Chapter 1 Indices

Solution: FLASHBACK CHAPTER 1

(a) (73 × 54)3 (b) (24 × 53 × 112)5 am × an = am + n
= 73(3) × 54(3) = 24(5) × 53(5) × 112(5) am ÷ an = am – n
= 79 × 512 = 220 × 515 × 1110 (am)n = amn

(c) (p2q3r)4 (d) (5m4n3)2 QU I Z
= p2(4) q3(4)r1(4) = 52m4(2)n3(2) mm = 256.
= p8q12r4 = 25m8n6 What is the value of m?

( ) ( ) (e) —2352 4 (f) —23yx–37 4 DISCUSSION CORNER

= —23 52((–44)) = —2344y–x7–3((–4 4)) Why is 1n = 1 for all
= —23 28–0 = —1816–yx–12– 28 values of n?
Discuss.

( g) —(36m— m2n—3n3) –3 (h ) —(2x—3y3—46)4x—1×0—y(132—xy2—)3

= —3 36m— m2(—33n)n1— 3(3 ) = —24 x—3(4—)y4—3(46)—x×10—3y31x—21(—3)y–2(–3)–

= 2—67mm—36n—n19 = —16—x12—3y616—x1×0—y2172—x3—y6

( ) = —29 m6 – 3 n9 – 1 = —16—3×6—27– x12 + 3 – 10y16 + 6 – 12

= —92 m3 n8 = 12x5 y10

MIND TEST 1.2e

1. Simplify each of the following.

(a) (2 × 34)2 (b) (113 × 95)3 (c) (133 ÷ 76)2 (d) (53 × 34)5

( ) ( ) ( e) ( m3n4 p2) 5 (f) (2 w 2 x 3)4 ( g ) —–b3—4a 5– 6 (h ) —32ba–45– 3

2. Simplify each of the following.

( ) (a) —11—13 1×–2 4—2 2 (b ) —33—×64—( 62—)3 ( ) (c) —6423– 3 ÷ —6432– ( d) —((–(—–4)4—6))62—××—((––—55)22—)3

( e) —x2—yx6y—×2 —x3 (f) —(h(h—3kk2)—2)4 (g) —((mm—52nn—37))2–3 ( h ) —((bb—22dd34—))23

3. Simplify each of the following.

( a) —(2m—2n1—42)m—3 7×n—1(23— m—n4)–2 (b) (—5x—y41—)52 x—×4y6—6x 1—0y (c) 2—(4dd—53ee—65) ×—× ((—36dd—e3e2–)43)–2 Saiz sebenar

13

How do you verify a0 = 1 and a–n = —a1n ; a ≠ 0? LEARNING
STANDARD
CHAPTER 1 Brainstorming 4 In pairs Verify that a0 = 1
and a–n = –a1–n ; a ≠ 0.
Aim: To determine the value of a number or an algebraic term with
a zero index.

Steps:
1. Study and complete the following table.
2. What is your conclusion regarding zero index?

Division in Solution Conclusion
index form from the
Law of indices Repeated multiplication solution

(a) 23 ÷ 23 23 – 3 = 20 —22 ××—22—××–22– = 1 20 = 1

(d) m5 ÷ m5 m5 – 5 = m0 —mm—×× —mm —×× mm—××—mm—××–mm–– = 1 m0 = 1

(c) 54 ÷ 54

(d) (–7)2 ÷ (–7)2

(e) n6 ÷ n6

Discussion:
2. What is your conclusion regarding zero index?

From Brainstorming 4, it is found that;

20 = 1
m0 = 1

Therefore, a number or an algebraic term with a zero index will give a value of 1.
In general, a0 = 1 ; a ≠ 0

H ow do you verify a–n = –a1–n – ?

Brainstorming 5 In groups

A im : T o v e rify a–n = —a1n.

Steps:

Saiz seb1e. naStrudy and complete the following table.

14

Chapter 1 Indices

Division in Law of indices Solution Conclusion CHAPTER 1
index form Repeated multiplication from the
solution

(a) 23 ÷ 25 23 – 5 = 2–2 —2 —× 22—××—22—×× 22––×––2 = –2–×1––2 = –21–2 2 –2 = 2–1–2
(b) m2 ÷ m5 m2 – 5 = m–3 –m––×—m—×m—m×—×m—m —× m— = —m —× m–1–×––m– = –m1–3 m –3 = –m–13–
(c) 32 ÷ 36

(d) (– 4)3 ÷ (– 4)7

(e) p4 ÷ p8

Discussion: Scan the QR Code or visit
2. What is your conclusion? Mathematics/F3/Chapter1
AlternativeMethod.mp4
From Brainstorming 5, it is found that; to watch a video that
describes alternative
2– 2 = 2—12 method to verify a–1 = —a1n.
m – 3 = m—13
BULLETIN
In general, a – n = –a1–n ; a ≠ 0
Negative index is a
Example 11 number or an algebraic
term that has an index of
1. State each of the following terms in positive index form. a negative value.
(a) a –2 (b) x – 4 (c ) –8–1–5–
(d ) y–1––9 – (e) 2 m –3 (f) —53 n –8 TIPS
♦ a –n = –a1–n
( ) ( ) ( g) –23– –10 (h ) –xy– –7
( ) ( )♦ ♦ a –nab – – n == – a–1––ba––n n

2. State each of the following in negative index form. REMINDER
2a –n ≠ 2—1an–
(a ) 3—14 (b) m—15 (c) 75
SMART MIND
( ) ( ) (d ) n20 (e) –54– 8 (f) –mn– 15
( ) – —94 –6 = x y
3. Simplify each of the following.
(a) 3 2 × 34 ÷ 38 (b) (—24—)2 —× (—35—)3 ( c) —(4x—y2—)2 —× x—5y What are the valuSeas iozf xsebenar
(28 × 36)2 (2x3y)5 and y?

15

CHAPTER 1 Solution:
1 . ( a ) a –2 = a–12– ( b) x – 4 = –x1–4 (c) –8–1–– 5 = 85 (d ) –y–1–– 9 = y9

( ) ( ) ( ) ( ) (e ) 2m –3 = –10 –7 –yx– 7
m­—23 (f ) —35 n– 8 = —53n– 8 ( g) –23– –32– 10 (h ) –xy–
= =

2 . ( a ) —314 = 3– 4 ( b) —m15 = m–5 (c) 75 = 7—1– 5 (d) n20 = n—–12– 0

( ) ( ) ( ) ( ) ( e) –45– 8= 15
–54– –8 (f ) –mn– –mn– –15
=

TIPS
3 . ( a ) 3 == = 233—3×1 –2 22+344 –÷ 8 38 (b) —(===2 22(—4—222)–828188—–6—××××1×6—(3—333×31635–51)—)3 22315 – 1 2 ( c) == —( 431——x4—(62y22x2—x2—x)2 325y2y—x4—×+)15×55—xy—– 5x5y15—5y1y4 + 1 – 5 yy01 == y1
= —2383 = —12 x–8 y0

= —21x–8

MIND TEST 1.2f

1. State each of the following terms in positive index form.

(a) 5–3 (b) 8– 4 (c) x– 8 (d) y–16 (e) —a1– –4

( ) ( f) —201–– –2 (g ) 3n– 4 (h ) – 5n– 6 ( i) —72 m–5 ( j) – —83 m– 4

( ) ( ) ( ) ( ) ( ) ­ (k) —25 –12 (l) – —37 –14 (m ) —yx –10 (n ) —23xy– – 4 (o ) —21x –5

2. State each of the following terms in negative index form.

(a) —514 (b) —813 (c) m—17 (d) —n19 (e) 102

( ) ( ) (f) (– 4)3 ( g) m12 (h ) n16 (i ) —47 9 (j ) —yx 10

3. Simplify each of the following.
( a) — (42(—)436—×)2 4–5 ( b ) —((22—3××—3342–))5–3 ( c) —(23—)(–52—2×)5–(5–4–)2

Saiz se be( nda) r— 3 m—2n9—4m×—3n(m5— n—3)––2 ( e ) —(2m—2n—2()9—–m3 —×3n()—32 m–n–2–)–4 ( f) (—2m—–2—(n4)m—5 2×—n(43)—2m–4n–)–2

16

Chapter 1 Indices

How do you determine and state the relationship between LEARNING CHAPTER 1
fractional indices and roots and powers? STANDARD

Relationship between n√a and a—n1 Determine and state the
relationship between
In Form 1, you have learnt about square and square root as well as cube fractional indices and
and cube root. Determine the value of x for roots and powers.

(a) x2 = 9 (b) x3 = 64 TIPS

Solution: ♦ 9 = 32 ♦ 64 = 43
Cube roots are used to
(a) x2 = 9 Square roots are used (b) x3 = 64 eliminate cubes.

√x2 = √32 to eliminate square s. 3√x3 = 3√43

x = 3 x = 4

Did you know that the values of x in examples (a) and (b) above can be determined by raising the
index to the power of its reciprocal?

(a) x2 = 9 The reciprocal (b) x3 = 64 BULLETIN
39—2 12 (—21 ) of 2 is —12 . oTfh3e risec—13ipr.oca l
x2(—21 )= x3(—31 )= 64(—31 ) —a1 is the reciprocal of a.
x1 = 43(—13 )
x1 = SMART MIND

x = 3 x = 4 What is the solution for
√– 4 ? Discuss.
From the two methods to determine the values of x in the examples
above, it is found that;

2√x = x–12
3√x = x–13

In general, n√a = a–1n ; a ≠ 0

Example 12
1. Convert each of the following terms into the form a—1n .

(a) 2√36 (b) 3√–27 (c) 5√m (d) 7√n

2. Convert each of the following terms into the form n√a . (d) n1—12
(a) 125— 15 (b) 256— 18 (c) (–1 000)— 31

3. Calculate the value of each of the following terms.

(a) 5√–32 (b) 6√729 (c) 512— 13 (d) (–243)—51

Solution:

1. (a) 2√36 = 36— 12 (b) 3√–27 = (–27)— 31 (c) 5√m = m— 51 (d) 7√n = n—17
2. (a) 125—51 = 5√125 (b) 25681– = 8√256
(c) (–1 000)—31 = 3√(–1 000) (d) n—112 = 12√n Saiz sebenar

17

CHAPTER 1 3. (a) 5√–32 = (–32)— 51 (b) 6√729 = 729—16 (c) 512 –31 = 83(–31) (d) (–243)—15 = (–3)5( —15 )

= (–2)5 (—51 ) = 36 (—61 ) = 81 = (–3)1
= 8 = –3
= (–2)1 = 31 TIPS

= –2 =3 You can use a scientific
calculator to check the

1. Convert each of the following terms into the form a–n1 . (d) 10√n

(a) 3√125 (b) 7√2 187 (c) 5√–1 024 (d) n–11–5

2. Convert each of the following terms into the form n√a. (d) (–32 768)—51

(a) 4—21 (b) 32—51 (c) (–729)—13

3. Calculate the value of each of the following terms.

(a) 3√343 (b) 5√–7 776 (c) 262 144—61

What is the relationship between a—mn and (am)—n1 , (a—n1 )m, n√am dan (n√a)m?

You have learnt that;

amn = (am)n and n√a1 = a—1n

From the two laws of indices above, we can convert a—mn into (am)—1n , (a—1n )m, n√am and (n√a)m.
Calculate the value of each of the following. Complete the table as shown in example (a).

a—mn (am)—n1 (a—n1 )m n√am (n√a)m
(3√64)2
(a) 64—32 (=64420)—9316(—31 ) (64—13 )2 3√642 = 42
= 163(—31 ) = 43(—31 )(2) = 3√4 096 = 16
= 42 = 16

= 16 = 16

(b) 16—43

(c) 243—52

Are your answers in (b) and (c) the same when you use different index forms? Discuss.

From the activity above, it is found that;

a—mn = (am)—1n = (a—­1n )m
a—mn = n√am = (n√a)m

Saiz sebenar

18

Chapter 1 Indices

Example 13

1. Convert each of the following into the form (am)–1n and (a–1n)m. CHAPTER 1

(a) 81­—23 (b) 27—32 (c) h—53
(c) m—25
2. Convert each of the following into the form n√am and (n√a)m.

(a) 343—32 (b) 4 096—65

Solution:

1. (a) 81­—23 = (813)—21 (b) 27—32 = (272)—31 (c) h—35 = (h3)—15
h—53 = (h—15 )3
81­—23 = (81—12 )3 27—23 = (27—13 )2

2. (a) 343—23 = 3√3432 (b) 4 096—56 = 6√4 0965 (c) m—52 = 5√m2

343—32 = (3√343)2 4 096—56 = (6√ 4 096)5 m—25 = (5√m)2

MIND TEST 1.2h
1. Complete the following table.

a—mn 729—65 121—32 w—37 ( ) ( )x—52—1816 —43 —hk —23

(am)—n1

(a—1n )m

n√ am

(n√ a )m

Example 14

1. Calculate the value of each of the following.

(a) 9—25 (b) 16—54

Solution:

1. (a) 9—52 (b) 16—54

Me thod 1 9—52 = (√9)5 = (3)5 = 243 Me thod 1 16—45 = (4√16)5 = 25 = 32

M etho d 2 9—25 = √95 = √59 049 = 243 Meth od 2 16—54 = 4√165 = 4√1 048 576 = 32

Saiz sebenar

19

MIND TEST 1.2i

CHAPTER 1 1. Calculate the value of each of the following.. (c) 128—27 (d) 256—38
(a) 27—32 (b) 32—25 (h) 49—32
(e) 64—34 (f) 1 024—52 (g) 1 296—43
(i) 2 401—14 (j) 121—32 (k) 2 197—23 (l) 10 000—43

2. Complete the following diagrams with correct values.

(a) � √6 561 � (b) 25�—3
5� 125�
27�—3 3�
243�—4 81 9� 125 625�—4
81� 3 125�—3

�√15 625�

How do you perform operations involving laws of LEARNING
indices? STANDARD

Law of indices Perform operations
involving laws of indices.
am × an = am + n a0 = 1 a–1n– = n√a
am ÷ an = am – n a–n = —a1n a–mn– = am(—n1 ) = (a—1n )m
(am)n = amn a–mn– = n√am = (n√a)m

Example 15

1. Simplify each of the following. ( c) ( —2h—)(28—×—13—(h1)6—–2h8—)—14–
(a) —(–3—x1)—03 8×—x4(2—yx33 —y––4–)2– ( b) —√m—(mn—–—143 —√×n(—3m)——n16 3)–—13–

Solution:

( a ) — ==(–( — —3—––1x230)———7)3183x×0xx——3—843×1(x×y——2—0434x8 2yx——3—x23y64xy–——3––y4(83–)2 –2 )–y — – 4–( 2–) ( b) —√= = m—— mm—n—— 1221———( —m43mmnn———–×—–—–34341—1 16×((———–√×61mn)mnnmn————3—41—323)31—)31—(——16–n– —6131)–n3(––1—13–) ( c) — ( = = 22 ——h2—2)2(h2h——288—22×3—— 3113(××——–13—((h)–1212)(——6—––)46h22h(——)(——4141 8h–—))h2h(—–)——8418–2((—)—4114––))

( )Saiz se be n ar== = –– —–1—2yx1x—71505–1 ×8y—–411 x3 + 6 – 4 y – 8 – 3 === mmmn—121—n32+ — 23—31 –(– –16)n —34 + 1 – — 41 == == 23—222252h2—+hh2– 1626×—–h(2––—212h) –h2 2 + 2 – (–2)

20

Example 16 Chapter 1 Indices

1. Calculate the value of each of the following. (c) (—24—3—54—×—5——32 )–2 CHAPTER 1
4√81 × √254
(a) ——49——12 ×—1—25—– —13—– ( b) —1(62——643 ×—×38—41)—–—12 — 14–
4√2 401 × 5√3 125 (c) (—24—3—54—×—5—23—)2
4√81 × √254
Solution: = —284—31——4154—(2×—) ×2–55–—42—–23–(–2)–
= —343(——415()——85×) ×—525–(—342–)
(a) ——49——12 ×—1—25—– ——31 ( b) —1(6—2—346—×× 83—14)–——21—14 –
4√2 401 × 5√3 125 = —3318—×× 5—534
= —224—6(3—(—3412—))××—3344—2(–(—21—–14) –) = 38 – 1 × 53 – 4
= —(77—24(—)12——14) ××—5(—53(5–)———3115 )– = 37 × 5–1
= —2233—××—33–2–1 = —357
= —7711—××—55–11– = —2 1—587–
= 71–1 × 5–1 –1 = 43 7 —25
= 70 × 5–2 = 23 – 3 × 3–1 – 2
= 1 × —512
= —215 = 20 × 3–3
= 1 × —313
= —217

MIND TEST 1.2j

1. Simplify each of the following.
(a) —3√—c2d(—c3–e—3d×—2ce—)13—2d 2—e—23– ( b) —(m—n(2m)—36×n—3()√——32m —n)–4 ( c) —√2—5√x33—y6zx—25y×—z48—x2z–

2. Calculate the value of each of the following..
( a ) — √479—–×4—×121—11 4 (b) —((15—2–35×—×37—62)9——13 ××—644√—)1–6–—13 – ( c) —4 √2—5(26—6××—√3—742×—95×2—)3—32√—12—5

(d) (—69 4—√)5—13—1×2—×(8—31√)——3344×—3 (×—14√–16–24–11– )–—41– (e) —(24—×—3166—)——3421 —××2—3√7—8—31 ×–—√8–1 (f) —64——23 ×—34√—21×—254√—×62—(25—× ——51 )––3

3. Given m = 2 and n = –3, calculate the value of 64—m3 × 512(– —1n ) ÷ 81—2nm . Saiz sebenar
4. Given a = —21 and b = —32 , calculate the value of 144a ÷ 64b × 256—ba .

21

How do you solve problems involving laws of indices? LEARNING
STANDARD
CHAPTER 1
Solve problems involving
laws of indices.

Example 17 FLASHBACK
Calculate the value of √3 × 12—23 ÷ 6 without using a calculator.
Common prime factors
of 6 and 12 are 2 and 3.

Understanding the Planning a strategy Implementing the strategy
problem
Convert each base √3 × 12—32 ÷ 6
Calculate the value of into prime factors and = 3—12 × (2 × 2 × 3)—32 ÷ (2 × 3)
numbers given in index calculate the value by = 3—12 × 2—32 × 2—23 × 3—32 ÷ (21 × 31)
form with different applying laws of indices. = 3—12 + —32 – 1 × 2—32 + —23 – 1
bases. = 31 × 22
= 12
Making a conclusion
√3 × 12—32 ÷ 6 = 12

Example 18 REMINDER
Calculate the value of x for the equation 3x × 9x + 5 ÷ 34 = 1.
♦ If am = an
Understanding the Planning a strategy then, m = n
problem
The question is an equation. ♦ If am = bm
Calculate the value of Hence, the value on the left side then, a = b
variable x which is part of the equation is the same as
of the indices indeks. the value on the right side of the Checking Answers
equation. Convert all the terms
into index form with base of 3. You can check the answer
by substituting the value of
x into the original equation.

3x × 9x + 5 ÷ 34 = 1

Left Right

Substitute x = –2 into left
side of the equation

Implementing the strategy Making a conclusion 3–2 × 9–2 + 5 ÷ 34

3x × 9x + 5 ÷ 34 = 1 3x + 6 = 0 If 3x × 9x + 5 ÷ 34 = 1, = 3–2 × 93 ÷ 34
then, x = –2
3x × 32(x + 5) ÷ 34 = 30 3x = – 6 = 3–2 × 32(3) ÷ 34

3x + 2(x + 5) – 4 = 30 x = —–36– = 3–2 + 6 – 4
3x + 2x + 10 – 4 = 30
x = –2 = 30 The same value
Saiz sebe nar 33x + 6 = 30 = 1 as the value on
the right side
amm== an of the equation.
n

22

Chapter 1 Indices

Calculate the possible values of x for the equation 3x2 × 32x = 315.
Substitute the values of x CHAPTER 1
Understanding Planning a Implementing the strategy into the original equation.
the problem strategy 3x2 × 32x = 315
3x2 × 32x = 315 If am = an,
Calculate All the 3x2 + 2x = 315 then, m = n. Left Right
the value of bases
x which is involved in x2 + 2x = 15 Solve the Substitute x = 3
part of the the equation quadratic
indices. are the Left: Right:
same.
x2 + 2x – 15 = 0 equation using 3(3)2 × 32(3) 315
factorisation
(x – 3)(x + 5) = 0 method. = 39 × 36

= 39 + 6

x – 3 = 0 or x + 5 = 0 = 315 The same value

x = 0 + 3 x = 0 – 5 Substitute x = –5

Making a conclusion x = 3 x = –5 Left: Right:

The possible values of x for 3(–5)2 × 32(–5) 315
the equation 3x2 × 32x= 315
are 3 and –5. = 325 × 3–10

= 325 + (–10)

= 315 The same value

Example 20 FLASHBACK

Solve the following simultaneous equations. Simultaneous linear
equations in two
2 5m × 5n = 58 and 2m × —21n = 2 variables can be solved
Solution: using substitution
2m × —21n = 2 method or elimination
25m × 5n = 58 method.
52(m) × 5n = 58

52m + n = 58 2m × 2–n = 21 Substitute m = 3 and n = 2
into original simultaneous
2m + n = 8 1 2m + (–n) = 21 equations.
m – n = 1 2
25m × 5n = 58
Equation  1 and 2 can be solved by substitution method.
From 1 : Left Right
2m + n = 8
Left: Right:
n = 8 – 2m 3 25m × 5n
= 52(m) × 5n 58
= 52(3) × 52
= 56 + 2
= 58 The same value
Substitute 3 into 2 Substitute m = 3 into 1
2m × —21n = 2
m – n = 1 2m + n = 8
2(3) + n = 8
m – (8 – 2m) = 1 6 + n = 8 You can also Left Right
substitute m = 3
m – 8 + 2m = 1 into equation Left: Right:
= 2 = m 22×33 ××—21 n—221–2 2
m + 2m = 1 + 8 n = 8 – 6 2 or 3 . 2

3m = 9 n = 2

m = —9 Hence, m = 3 and n = 2. = 23 + (–2)
3
= 21
m = 3 = 2 The Saizsame value sebenar

23

Example 21

CHAPTER 1 My equation is
3(9x) = 27y.

My equation is
16(4x) = 16 y.

The values of the variables x
and y can be determined if you
can solve both the equations.

Chong and Navin performed an experiment to determine the relationship between variable x and
variable y. The equation Chong obtained was 16(4x) = 16 y, while the equation Navin got was
3(9x) = 27y as the findings of the experiment they performed. Calculate the value of x and of y
which satisfy both the experiments Chong and Navin have performed.

Solution:

16(4x) = 16y 3(9x) = 27y You can also substitute
42(4x) = 42(y) 3(32x) = 33(y) y = 3 into equation
42 + x = 42y 31 + 2x = 33y 2 or 3 .

2 + x = 2y 1 1 + 2x = 3y 2

Equations 1 and 2 can be solved by elimination Substitute y = 3 into equation 1

method. Multiply equation 1 1 : 2 + x = 2y
1 × 2 : 4 + 2x = 4y by 2 to equate the 2 + x = 2(3)
x = 6 – 2
3 coefficients of variable x. x = 4

2 : 1 + 2x = 3y Hence, x = 4, y = 3

3 – 2 :

3 + 0 = y

y = 3

Dynamic Challenge

Test Yourself

1. State whether each of the following operations which involves the laws of indices is true or
false. If it is false, state the correct answer.

(a) a5 = a × a × a × a × a (b) 52 = 10 (c) 30 = 0
(f) 2a– 4 = —21a–4
(d) (2x3)5 = 2x15 (e) m0n0 = 1 (i) (5m—14 )– 4 = 6—m25–

Saiz se ben(ga) r32—52 = (2√32)5 ( ) ( )(h) —mn –4 —mn 4

=

24

Chapter 1 Indices

2. Copy and complete the following diagram with suitable values.

5□ × 55 53(□) CHAPTER 1

( ) —51□ 3 512 ÷ 5□

5—1□ 59 (√25)□

— 56 —5×25–□– ( ) —15 □
(5□)—32
(□√125)□

3. Copy and complete the following diagram.

Operations that ( )20 as –31—– 4 as —53 –2 as 72 × 5–3 as (5–1 × √25)3
involve laws
of indices
Value

Skills Enhancement (c) √xy × 3√xy2 × 6√xy5

1. Simplify each of the following. (c) (256)—38 × 2–3
(a) (m n4)3 ÷ m 4n5 (b) 3x × —16 y4 × (xy)3 (f) (125)—32 × (25)– —32 ÷ (625)– —41

2. Calculate the value of each of the following. (c) axa8 = 1
(f) 2x = —12610–x
(a) 64—31 × 5–3 (b) 7–1 × 125—23 (i) 25x ÷ 125 = —51x

(d) 24 × 16– —43 (e) √49 × 3–2 ÷ (√81)–1 Saiz sebenar

3. Calculate the value of x for each of the following equations. 25

(a) 26 ÷ 2x = 8 (b) 3– 4 × 81 = 3x

(d) 4 × 8x + 1 = 22x (e) (ax)2 × a5 = a3x

(g) 3 6 ÷ 3x = 8 1(x – 1) (h) (m2)x × m(x + 1) = m–2

Self Mastery

CHAPTER 1 1. Calculate the value of each of the following without using a calculator.

(a) 4—13 × 50—32 × 10—53 (b) 5—52 × 20—32 ÷ 10–2 (c) 60—12 × 125—32 ÷ √15

2. Calculate the value of x for each of the following equations.
(a) 6 4x­—12 = 27 x– —25 (b) 3x—23 = —247 x– —34 (c) 25x– —32 – —35 x—31 = 0

3. Calculate the possible values of x for each of the following equations.

(a) ax2 ÷ a5x = a6 (b) 2x2 × 26x = 27 (c) 5x2 ÷ 53x = 625

4. Solve the following simultaneous equations. (b) 4(4x) = 8y + 2 and 9x × 27y = 1
(a) 81(x + 1) × 9x = 35 and 82x × 4(22y) = 128

5. In an experiment performed by Susan, it was found

that the temperature of a metal rose from 25˚C to T˚C
according to equation T = 25(1.2)m when the metal

was heated for m seconds. Calculate the difference in

temperature between the fifth second and the sixth

second, to the nearest degree Celsius.

6. Encik Azmi bought a locally made car for RM55 000.

After 6 years, Encik Azmi wishes to sell the car. Based

on the explanation from the used car buyers, the price RM55 000

of Encik Azmi’s car will be calculated by the formula

( )RM55000 —89 n

. In this situation, n is the number of years

after the car is bought. What is the market value of Encik

7. Mrs Kiran Kaur saved RM50 000 on 1 March 2019
in a local bank with an interest of 3.5% per annum.
After t years, Mrs Kiran Kaur’s total savings, in RM,
is 50 000 (1.035)t. Calculate her total savings on

1 March 2025, if Mrs Kiran Kaur does not withdraw
her savings.

Saiz sebenar

26

Chapter 1 Indices

P ROJ EC T

Materials: One sheet of A4 paper, a pair of scissors, a long ruler, a pencil. CHAPTER 1

Instructions: (a) Carry out the project in small groups.
(b) Cut the A4 paper into the shape of a square. (Biggest possible)
Steps:

1. Draw the axes of symmetry (vertical and horizontal only) as shown in Diagram 1.

2. Calculate the number of squares formed. Write your answers in the space provided in
Sheet A.

3. Draw the vertical and horizontal axes of symmetry for each square as shown in
Diagram 2.

4. Calculate the number of squares formed. Write your answers in the space provided in
Sheet A.

5. Repeat step 3 and step 4 as many times as possible. 8

11

2 7
2 6

3

Diagram 1 45
Diagram 2

6. Compare your answers with those of other groups. Scan the QR Code or
7. What can you say about the patterns in the column ‘Index form’ visit http://bukutekskssm.
in Sheet A? my/Mathematics/F3/
8. Discuss the patterns you identify. Chapter1SheetA.pdf
Sheet A

Number of axes Index form Number of Index form
of symmetry squares 20
– 22
0 21 1
2 4 Saiz sebenar
8 16

27

CHAPTER 1 an Index CONCEPT MAP 54 = 5 × 5 × 5 × 5
Base m × m × m × m × m = m5
Indices
an = a × a × a × … × a

n factors

Multiplication Division Power
am × an = am + n am ÷ an = am – n (am)n = amn (am × an)p = amp × anp
(34)2= 38 (3a4)3 = 27a12
23 × 25 = 23 + 5 36 ÷ 34 = 36 – 4

Fractional index Negative index Zero index
a 5 ––n3 == —­—a511n3 ; a ≠ 0 a0 = 1 ; a ≠ 0
a—­ 1n = n√a 8–13 = 3√8 20 = 1
a—mn = (am)—n1 = (a—n1 )m 8–32 = (82)–13 = (8–13)2 m0 = 1
a—mn = n√am = (n√a)m 8–32 = 3√82 = (3√8)2

SELF-REFLECT

At the end of this chapter, I can:

1. Represent repeated multiplication in index form and describe its meaning.

2. Rewrite a number in index form and vice versa.

3. Relate the multiplication of numbers in index form with the same base, to repeated
multiplications, and hence make generalisation.

4. Relate the division of numbers in index form with the same base, to repeated
multiplications, and hence make generalisation.

5. Relate the numbers in index form raised to a power, to repeated multiplication, and hence
make generalisation.

6. Verify that a0 = 1 and a–n = a—1n ; a ≠ 0.

7. Determine and state the relationship between fractional indices and roots and powers.

8. Perform operations involving laws of indices.

Saiz seb9.enSaorlve problems involving laws of indices.

28

Chapter 1 Indices

EXPLORING MATHEMATICS

Do you still remember the Pascal’s Triangle that you learnt in the Chapter 1 Patterns and CHAPTER 1
Sequences in Form 2?

The Pascal’s Triangle, invented by a French mathematician, Blaise Pascal, has a lot of unique
properties. Let us explore two unique properties found in the Pascal’s Triangle.

Activity 1 Sum Index form

1 1 20
11 2 21
121 4 22

1331

14641

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1

1 10 45 120 210 252 210 120 45 10 1

Instructions: Sheet 1 Sheet 1(a)

1. Carry out the activity in pairs.

2. Construct the Pascal’s Triangle as in Sheet 1.

3. Calculate the sum of the numbers in each row. Write the sum in index form with base of 2.

Activity 2 115 = 161 051
1 5 10 10 5 1
11n Value
110 1 1 +1 +—111
111 11 11 161051
112 121
113 1 331 121
114 1331
115 14641
116 1 5 10 10 5 1
117 1 6 15 20 15 6 1
118 1 7 21 35 35 21 7 1
119 1 8 28 56 70 56 28 8 1
1110 1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
Sheet 2(a)
Sheet 2

Instructions:

1. Carry out the activity in small groups.

2. Construct the Pascal’s Triangle as in Sheet 2.

3. Take note on the numbers in each row. Each number is the value of index with base of 11.

4. Complete Sheet 2(a) with the value of index with base of 11 without using a calculator.

5. Present your results. Saiz sebenar