Mathematics_Form_3 Textbook

DUAL LANGUAGE PROGRAMME

FORM

RUKUN NEGARA

Bahawasanya Negara Kita Malaysia
mendukung cita-cita hendak;

Mencapai perpaduan yang lebih erat dalam kalangan
seluruh masyarakatnya;

Memelihara satu cara hidup demokrasi;

Mencipta satu masyarakat yang adil di mana kemakmuran negara
akan dapat dinikmati bersama secara adil dan saksama;

Menjamin satu cara yang liberal terhadap
tradisi-tradisi kebudayaannya yang kaya dan pelbagai corak;

Membina satu masyarakat progresif yang akan menggunakan
sains dan teknologi moden;

MAKA KAMI, rakyat Malaysia,
berikrar akan menumpukan

seluruh tenaga dan usaha kami untuk mencapai cita-cita tersebut
berdasarkan prinsip-prinsip yang berikut:

KEPERCAYAAN KEPADA TUHAN
KESETIAAN KEPADA RAJA DAN NEGARA

KELUHURAN DAN PERLEMBAGAAN
KEDAULATAN UNDANG-UNDANG
KESOPANAN DAN KESUSILAAN

(Sumber : Jabatan Penerangan, Kementerian Komunikasi dan Multimedia Malaysia)

KURIKULUM STANDARD SEKOLAH MENENGAH

MATHEMATICS

FORM 3

Authors
Chiu Kam Choon
Vincent De Selva A/L Santhanasamy
Punithah Krishnan
Raja Devi Raja Gopal

Translator
Yew Chian-Hauo

Editors
Premah A/P Rasamanie
Muhammad Amirullah Bin Miswan

Lai Boon Sing

Designers
Lim Fay Lee
Nur Syahidah Mohd Sharif

Illustrators
Asparizal Mohamed Sudin
Mohammad Kamal B Ahmad

Penerbitan Pelangi Sdn Bhd.

2019

KEMENTERIAN
PENDIDIKAN
MALAYSIA

Book Series No: 0166 ACKNOWLEDGEMENTS

KPM2019 ISBN 978-983-00-9651-3 The publishing of this textbook involves
First Published 2019 cooperation from various parties. Our
© Ministry of Education Malaysia wholehearted appreciation and gratitude goes
out to all involved parties:
All rights reserved. No part of this book may
be reproduced, stored in a retrieval system, ‡ &RPPLWWHH PHPEHUV RI Penambahbaikan
or transmitted in any form or by any means, Pruf Muka Surat, Textbook Division,
either electronic, mechanical, photocopying, Ministry of Education, Malaysia.
recording or otherwise, without the prior
permission of Director General of Education ‡ &RPPLWWHH PHPEHUV RI Penyemakan
Malaysia, Ministry of Education Malaysia. Pembetulan Pruf Muka Surat, Textbook
Negotiations are subject to an estimation of Division, Ministry of Education, Malaysia.
royalty or an honorarium.
‡ &RPPLWWHH PHPEHUV RI Penyemakan
Published for the Ministry of Education Naskhah Sedia Kamera, Textbook Division,
Malaysia by: Ministry of Education, Malaysia.
PENERBITAN PELANGI SDN. BHD.
66, Jalan Pingai, Taman Pelangi, ‡ 2൶FHUV LQ 7H[WERRN 'LYLVLRQ DQG WKH
80400 Johor Bahru, Curriculum Development Division,
Johor Darul Takzim. Ministry of Education, Malaysia.

Layout and Typesetting: ‡ &KDLUSHUVRQ DQG PHPEHUV RI WKH 4XDOLW\
PENERBITAN PELANGI SDN. BHD. Control Panel.
Font type: Times New Roman
Font size: 11 point ‡ (QJOLVK /DQJXDJH 7HDFKLQJ &HQWUH (/7&
Teacher Education Division, Ministry of
Printed by: Education Malaysia.
THE COMERCIAL PRESS SDN. BHD.
Lot 8, Jalan P10/10, ‡ (GLWRULDO 7HDP DQG 3URGXFWLRQ 7HDP
Kawasan Perusahaan Bangi, especially the illustrators and designers.
Bandar Baru Bangi,
43650 Bangi, ‡ (YHU\RQH ZKR KDV EHHQ GLUHFWO\ RU LQGLUHFWO\
Selangor Darul Ehsan. involved in the successful publication of
this book.

ii

Introduction v

Symbols and Formulae vii

1CHAPTER Indices 1
1.1 Index Notation 2
1.2 Law of Indices 6

2CHAPTER Standard Form 30
6LJQL¿FDQW )LJXUHV
2.2 Standard Form 37

3CHAPTER Consumer Mathematics: Savings and Investments, 50
Credit and Debt 52
73
3.1 Savings and Investments

3.2 Credit and Debt Management

4CHAPTER Scale Drawings 86
4.1 Scale Drawings 88

5CHAPTER Trigonometric Ratios 106
108
5.1 Sine, Cosine and Tangent of Acute Angles in Right-angled
Triangles

iii

6CHAPTER Angles and Tangents of Circles 128
6.1 Angle at the Circumference and Central Angle Subtended
130
by an Arc
150
&\FOLF 4XDGULODWHUDOV 160
6.3 Tangents to Circles
6.4 Angles and Tangents of Circles

7CHAPTER Plans and Elevations 168
7.1 Orthogonal Projections 170
7.2 Plans and Elevations 182

8CHAPTER Loci in Two Dimensions 198
8.1 Loci 200
8.2 Loci in Two Dimensions 204

9CHAPTER Straight Lines 224
9.1 Straight Lines 226

Answers 252
Glossary 262
References 263
Index 264

iv

Introduction

This Form 3 Mathematics Textbook is prepared based on Kurikulum Standard Sekolah
Menengah (KSSM). This book contains 9 chapters arranged systematically based on
Form 3 Mathematics Dokumen Standard Kurikulum dan Pentaksiran (DSKP).

At the beginning of each chapter, pupils are introduced to materials related to daily
life to stimulate their thinking about the content. The Learning Standard and word lists are
included to provide a visual summary of the chapter’s content.

Special features of this book are: Description

What will you learn? Contains learning standards that pupils need to
WWhhyy ddoo yyoouu lleeaarrnn tthhiiss cchhaapptteerr?? achieve in each chapter.

$SSOLFDWLRQV RI NQRZOHGJH LQ UHODWHG FDUHHU ¿HOGV

EExxpplloorriinngg EErraa Historical background or origin of the content.
Word list contained in each chapter.
WORD B A N K Helps pupils to understand the basic mathematical
concept via individual, pair or group activities.
Brainstorming Provides additional information about the content
learned.
Individual In pairs In groups

BULLETIN

QU I Z Questions that test pupils’ ability to understand
basic concepts in each chapter.

REMINDER Additional facts that pupils need to be reminded of
TIPS and common mistakes to be avoided.

SMART MIND Exposes pupils to additional knowledge that they
need to know.
Challenging tasks for enhancement of critical and
creative thinking skills.

v

SMART Description
DISCUSSION CORNER
Exposes pupils to the use of technology in
FLASHBACK mathematics.
SMART FINGER 1,234567.89 Develops pupils’ mathematical communication
skills.
7 8 9÷ Helps pupils to recall what they have learnt.
4 5 6x
1 2 3- 6KRZV KRZ WR XVH VFLHQWL¿F FDOFXODWRUV
AC 0 . + Enables pupils to carry out and present project work.
Assesses pupils’ understanding on the concepts they
P ROJ EC T have learnt.
4XHVWLRQV WR HQKDQFH SXSLOV¶ KLJKHU RUGHU WKLQNLQJ
MIND TEST skills.
3URYLGHV GLYHUVL¿HG WDVNV ZKLFK LQFRUSRUDWH WKH
Dynamic Challenge elements of LOTS, HOTS, TIMSS and PISA.

EXPLORING MATHEMATICS (QDEOHV SXSLOV WR VFDQ D 45 &RGH XVLQJ D PRELOH
device to access further information.
CONCEPT MAP
Covers applicable concepts of digital tool calculators,
SELF-REFLECT KDQGV RQ DFWLYLWLHV DQG JDPHV WKDW DLP WR H൵HFWLYHO\
enhance pupils’ understanding.
Checking Answers
Overall chapter summary.
S T EM
Pupils self-assess their achievement.

Checking answers using alternative methods.
Activities with elements of Science, Technology,
Engineering and Mathematics.

vi

Symbols and Formulae

SYMBOLS

¥ root у is more than or equal to
ʌ SL Ͻ is less than
a : b ratio of a to b р is less than or equal to

A × 10n standard form where 6 WULDQJOH
1 р A Ͻ 10 and n LV DQ LQWHJHU
 DQJOH
= is equal to
$ GHJUHH
§ is approximately equal to ' minute
 is not equal to '' second
Ͼ is more than

FORMULAE tan θ = —scion—sθθ–

am × an = am + n 3\WKDJRUDV WKHRUHP:
am ÷ an = am – n
(am)n = amn cb c2 = a2 + b2
a0 = 1 b2 = c2 – a2
a–n = —a1n
a—n1 = n¥ a a a2 = c2 – b2
a—mn = (am)—n1 = (a—n1 )m
Distance between ¥ x2 – x1)2 + (y2 – y1)2
a—mn = n¥ am = (n¥ a )m
I = Prt two points
MV = P(1 + —nr )nt
A = P + Prt ( )Midpoint = ²x1²+2 ²x2 ²y1²+2 ²y2

sin θ = —oh—pypp—oosti—eten—usisd—ee – hypotenuse *UDGLHQW m = —hov—reirz—toicn—atla—ldids—itsatna—cnec—e
cos θ = —ahd—yjap—coet—nent u—sisde—e – opposite side
tan θ = —oa—dpjpa—ocsei—ntet —ssiidd—ee – ș m = —xy22—–– yx—11
adjacent m = – —yx--—iinntt—eerr—cceepp—tt

side

http://bukutekskssm. Download a free QR Code scanner application to your mobile device. Scan QR Code
my/Mathematics/F3/ or visit the website http://bukutekskssm.my/Mathematics/F3/Index.html to download
¿OHV IRU EUDLQVWRUPLQJ 7KHQ VDYH WKH GRZQORDGHG ¿OH IRU R൷LQH XVH
Index.html Note: Students can download the free GeoGebra and Geometer’s Sketchpad
*63 VRIWZDUH WR RSHQ UHODWHG ¿OHV

vii

1CHAPTER Indices

What will you learn?

1.1 Index Notation

1.2 Law of Indices

WWhhyy ddoo yyoouu lleeaarrnn tthhiiss cchhaapptteerr??
‡ :ULWLQJ D QXPEHU LQ LQGH[ QRWDWLRQ HQDEOHV WKH

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.
‡ &RQFHSW RI LQGH[ LV XVHG LQ WKH ILHOGV RI VFLHQFH
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 metres. 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, a well-known British mathematician,
GHYHORSHG WKH ¿HOG RI LQGH[ QRWDWLRQ DQG LQWURGXFHG
negative indices and fractional indices.

http://bukutekskssm.my/Mathematics/F3/
([SORULQJ(UD&KDSWHU SGI

WORD B A N K

‡ EDVH ‡ DVDV

‡ IDFWRU ‡ IDNWRU

‡ LQGH[ ‡ LQGHNV

‡ IUDFWLRQDO LQGH[ ‡ LQGHNV SHFDKDQ

‡ SRZHU ‡ NXDVD

‡ URRW ‡ SXQFD NXDVD

‡ LQGH[ QRWDWLRQ ‡ WDWDWDQGD LQGHNV

1

1.1 Index Notation

CHAPTER 1 What is repeated multiplication in index form? LEARNING

The development of technology not only makes most of our daily STANDARD

WDVNV HDVLHU LW DOVR VDYHV H[SHQVHV LQ YDULRXV ¿HOGV )RU LQVWDQFH Represent repeated
multiplication in index form
the use of memory cards in digital cameras enable users to store and describe its meaning.

photographs in a large number and to delete or edit unsuitable

photographs before printing. DISCUSSION CORNER

Discuss the value of the
capacity of a pen drive.

BULLETIN

7KH QXFOHDU ¿VVLRQ RI
uranium U-320 follows the
pattern 30, 31, 32, …

In the early stage, memory cards were made with a capacity of 4MB. The capacity increases
over time to meet the demands of users. Do you know that the capacity of memory cards is
calculated using a special form that is 2n?

,Q )RUP \RX KDYH OHDUQW WKDW 3 = 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 î î )RU H[DPSOH

4×4=42 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=43
The value of index is 3
Repeated 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
(c) ( –2) × (–2) × (–2) ( d) — 4 × —4 × — 4 × — 4 × —4
(e) m × m × m × m × m × m × m (f) n × n × n × n × n × n × n × n

2

Chapter 1 Indices

Solution: (b) 0.3 × 0.3 × 0.3 × 0.3 = (0.3)4 CHAPTER 1
(a) 5 × 5 × 5 × 5 × 5 × 5 = 56
repeated four times
repeated six times
( ) (d) — × — × — × — × — = — 5
(c) ( –2) × (–2) × (–2) = (–2)3 4 44 44 4
UHSHDWHG ¿YH WLPHV
repeated three times (f) n × n × n × n × n × n × n × n = n8

(e) m × m × m × m × m × m × m = m7 repeated eight times

repeated seven times

)URP WKH VROXWLRQ LQ ([DPSOH LW LV IRXQG WKDW WKH YDOXH RI LQGH[ LQ DQ LQGH[ IRUP LV WKH VDPH DV
the number of times the base is multiplied repeatedly. In general,

an = a × a × a × … × a ; a 
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
( ) ( ) —2 m6– —3 4—2
7 n 6
9
n0 (0.2)9 x 4
8 2
( )x20 2 — 3 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) — 2 × — 2 × — 2 × — 2 (d) (–m) × (–m) × (–m) × (–m) × (–m)
H ²3 î ²3 î ²3
( ) ( ) ( ) ( ) ( ) ( ) I – –n × – n– × – – n × – n– × – – n × – n–

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

( a) ( –3)3 ( b) ( 2.5)4 ( )( c) —3 5 ( )( d) – 2 —4 3

(e) k6 ( f) ( –p)7 ( )( g) —m 8 (h) (3n)5

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
‡ LV GLYLGHG UHSHDWHGO\ ‡ LV GLYLGHG ‡ LV GLYLGHG

by 2. repeatedly by 4. repeatedly by 8.

2 ) 64 4 ) 64 8 ) 64
n=2 8) 8
2 ) 32 n=3 4) 4

n=6
2) 8 Hence, 64 = 82

2) 4

2) 2 Hence, 64 = 43

The division is
Hence, 64 = 26
continued until
LV REWDLQHG

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 64 Hence, 64 = 82
8
Hence, 64 = 43 DISCUSSION CORNER

32 Which method is easier
64 to convert a number into
a number in index form?
Hence, 64 = 26 Is it the repeated division
or repeated multiplication
method? Discuss.

4

Example 3 Chapter 1 Indices CHAPTER 1

Write — 3 —2 – in index form using base of —2 . Repeated Multiplication Method
Solution: —25 × —25 × —25 × —25 × —52
—245–
Repeated Division Method —8 –
—6 2 –5
n=5 2 ) 32 n=5 — 3 —2 –
5 ) 625
2) 8 ( )Hence, — 3 —2 – = — 2 5
2) 4 5 ) 25
2) 2 5) 5


( )Hence, — 3 —2 – = — 2 5

MIND TEST 1.1b

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

D [base of 3] E [base of 5] ( c) — 6 4– [ ]base of —45

base of – —
4
[ ( )] ( d) 0 .00032 [base of 0.2] H ± [base of (– 4)] ( f) —

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

The value of an FDQ EH GHWHUPLQHG E\ UHSHDWHG PXOWLSOLFDWLRQ PHWKRG RU XVLQJ D VFLHQWL¿F
calculator.

Example 4

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

(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

î
î 3

32 Hence, 0.63
Hence, 25 = 32

5

Example 5 SMART FINGER 1,234567.89 REMINDER
7 8 9÷
CHAPTER 1 (a) 54 = 625 4 5 6x = Negative or fractional base
(b) (–7)3 = –343 1 2 3- 4= must be placed within
AC 0 . + brackets when using a
( ) ( c) — 4 — – 5 ^4= )^ calculator to calculate
( (–) 7 ) ^ 3 ^6 values of given numbers.
=
( 2 ab/c 3 ) ^ DISCUSSION CORNER
( )(d) ²35 2 —6254– 2=
( 1 ab/c 3 ab/c 5 = Calculate questions (c),
= (d) and (e) in Example 5
( (–) 0 . 5 ) without using brackets.
(e) (– 0.5)6 Are the answers the
same? Discuss.
MIND TEST 1.1c

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

(a) 94 (b) (– 4)5 (c) (2.5)3 (d) (– 3.2)3

( )( e) — 8 5 ( )( f) – — 6 4 ( )( g) ² 3 2 ( ) (h) – 2 — 3 3

1.2 Law of Indices

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

multiplication.

Steps:
1. Study example (a) and complete examples (b) and (c).
2. Discuss with your friend and state three other examples.
3. ([KLELW WKUHH H[DPSOHV LQ WKH PDWKHPDWLFV FRUQHU IRU RWKHU JURXSV WR JLYH IHHGEDFN

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)

(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?

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

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 × 73 (b) (0.2)2 × (0.2)4 × (0.2)5 ( c) 2 k2 × 4k3 ( d) 3 m4 × —6 m5 î m

Solution: (b) (0.2)2 × (0.2)4 × (0.2)5 REMINDER
= (0.2)2 + 4 + 5 a = a1
(a) 72 × 73 = (0.2)
= 72 + 3
= 75

( c) 2k2 × 4k3 (d) 3m4 × — m5 î m
= (2 × 4)(k2 × k3) 6

Operation of î ² î m4 × m5 × m )
WKH FRHI¿FLHQWV 6

= 8k2 + 3 = 6m SMART MIND
= 8k5 = 6m If ma × mb = m8, such

that a > 0 and b > 0,

what are the possible

MIND TEST 1.2a values of a and b?

1. Simplify each of the following. (b) (– 0.4)4 × (– 0.4)3 × (– 0.4)
(a) 32 × 3 × 34
( ) ( ) ( )(d) ± ²52 2 × ± ²52 3 × ± ²52 5
( ) ( ) ( )(c) —4 × —4 3 × —4 5
77 7 ( f) n 6 × — n2 × — n3 × n
( e) 4m2 × — m3 × (– 3)m4 25 4

2 (h) – — y5 × (– 6)y3 × — y4
(g) – x4 × — x × — x2 23

45

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
VDPH EDVH ¿UVW 7KHQ DGG
the indices for the terms
with the same base.

Simplify each of the following.

(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 × — 4 n2

Solution:

(a) m3 × n2 × m4 × n5 (b) (0.3)2 × (0.2)2 × 0.3 × (0.2)5 × (0.3)3
= (0.3)2 × (0.3) × (0.3)3 × (0.2)2 × (0.2)5
= m3 × m4 × n2 × n5 Group the terms = (0.3) × (0.2)(2 + 5)
= m3 + 4 × n2 + 5 with the same base. = (0.3)6 × (0.2)7

= m7 × n7 Add the indices for terms
= m7n7 with the same base.

( c) p2 × m3 × p4 × n3 × m4 × n2 (d) –m4 × 2n5 × 3m × —4 n2

= m3 × m4 × n3 × n2 × p2 × p4 ± î î î ² 4 m4 × m × n5 × n2
= m3 + 4 × n3 + 2 × p2 + 4
= m7 n5 p6 = – —23 m n5 + 2 REMINDER

= – —23 m5 n7 –an  ±a)n

Example:
± 2  ± 2

MIND TEST 1.2b ± 

1. State in the simplest index form. (b) (0.4)2 î 3 î î 5 î
(a) 54 × 93 × 5 × 92 ( d) –2k5 × p6 × — 4 p5 × 3k

F x5 × y3 × — 2 x × — 3 y4

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).
2. Discuss with your friend and state three other examples.
3. 3UHVHQW \RXU ¿QGLQJV

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 = 43
(b) 26 ÷ 22 4×4
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?

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW 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) m ÷ 4m5 ÷ m2 I ± p8 ÷ 2p5 ÷ 4p2

Solution: (c) m4n3 ÷ m2n
= m4n3 ÷ m2n
(a) 54 ÷ 52 (b) (–3)4 ÷ (–3)2 ÷ (–3) = m4 – 2 n ±
= (–3)4 ÷ (–3)2 ÷ (–3) = m2 n2
= 54 – 2 = (–3) ± ±
= 52 = (–3)

= –3

9

(d) 25x2y3 ÷ 5xy H m ÷ 4m5 ÷ m2 I ± p8 ÷ 2p5 ÷ 4p2
= — 4 (m ÷ m 5 ÷ m2) = ±—2 – (p8 ÷ p5) ÷ 4p2
CHAPTER 1 = 2 5 x2y3 ÷ 5x y
= 3(m ± ) ÷ m2 = –8p8–5 ÷ 4p2
= —255 x ± y ± = –8p3 ÷ 4p2
Operation of the = 3m5 – 2
= 5x y2 FRHI¿FLHQWV = – —84 (p3 ÷ p2)
= 3m3 = –2p3 – 2
= 5xy2 = –2p

= –2p

MIND TEST 1.2c

1. Simplify each of the following.

(a) 45 ÷ 44 (b) 7 ÷ 76 ÷ 72 (c) —mm84—nn6
(f) –25h4 ÷ 5h2 ÷ h
(d) —297x—x34y–y2–5 (e) m7 ÷ m2 ÷ m4

2. Copy and complete each of the following equations.

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

(c) —m — n—m4 ×7—nm——n2= m5n (d) —27—x3y—6x×2—yx3—y – = 3x y5

3. If —224x—×× —332y = 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.
3. 3UHVHQW \RXU ¿QGLQJV

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

= 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 = 5 = 4

32(4) = 32 × 4 54(3) = 54 × 3 43(6) = 43 × 6
= 38 = 5 = 4

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW SMART MIND

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

Example 9

1. Simplify each of the following.

(a) (34)2 (b) (h3) (c) ((–y)6)3

2. Determine whether the following equations are true or false.

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

11

CHAPTER 1 Solution: (b) (h3) (c) ((–y)6)3
= (–y)6(3)
1. (a) (34)2 = h = (–y)
= h30
= 34(2) (c) (32)6 = (272)4
= 38 (b) (23)4 = (22)6
Same left right
2. (a) (42)3 = (43)2 left right
Same /HIW
left right /HIW (32)6 = 32(6) = 3
(23)4 = 23(4) = 2
/HIW 5LJKW
(42)3 = 42(3) = 46 5LJKW (272)4 = (33(2))4 Not the
(22)6 = 22(6) = 2
5LJKW same
(43)2 = 43(2) = 46 Hence, (23)4 = (22)6
is true. = 36(4)
Hence, (42)3= (43)2 = 324
is true. Hence, (32)6 = (272)4
is false.

MIND TEST 1.2d

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

D 5)2 (b) (3 )2 (c) (72)3 (d) ((– 4)3)7
(e) (k8)3 (f) (g2) (g) ((–m)4)3 (h) ((–c)7)3

2. Determine whether the following equations are true or false. (d) – (72)4 = (– 492)3

(a) (24)5 = (22) (b) (33)7 = (272)4 (c) (52)5 2)3

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

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

(am ÷ bn)q
= (am)q ÷ (bn)q
= amq ÷ bnq

Example 10

1. Simplify each of the following.

(a) (73 × 54)3 (b) (24 × 53 î 2)5 (c) (p2q3r)4 (d) (5m4n3)2
(g) —(36m—m2n—3n3)–3 (h) —(2x—3y3—46)4x— × y—( 3 —xy2—)3
( )(e) —3252 4 ( )(f) —32xy–73 4

12

Chapter 1 Indices

Solution: (b) (24 × 53 î 2)5 FLASHBACK CHAPTER 1
= 24(5) × 53(5) î 2(5)
(a) (73 × 54)3 = 220 × 5 î am × an = am + n
= 73(3) × 54(3) am ÷ an = am – n
= 79 × 5 (am)n = amn

(c) (p2q3r)4 (d) (5m4n3)2 QU I Z
= p2(4) q3(4)r = 52m4(2)n3(2) mm = 256.
= p8q r4 = 25m8n6 What is the value of m?

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

= —3252((–44)) = —2344y–x7–3((–44)) Why is 1n = 1 for all
= —2328–0 = — –yx–2 – 8 values of n?
Discuss.

(g) —(3m—2n—3)–3 (h) —(2x—3y3—46)4x— × —y( 3 —xy2—)3
6m3n = — 24x—3 ( 4 — ) y 4 —3(46)—x× —3y3 x— — y–2(–3)–
= — —x —3y6 —x × —y2 7 —x3—y6
= —336m—m2(—33n)n — 3(3)
( ) = — — î— – x ± y ±
= 2—67mm—36n—n 9 36
x5 y
= — m6 – 3 n ±
2

= —2 m3 n8

MIND TEST 1.2e

1. Simplify each of the following.

(a) (2 × 34)2 E 3 × 95)3 F 3 ÷ 76)2 (d) (53 × 34)5

(e) (m3n4p2)5 (f) (2w2 x 3)4 ( )(g) —–b3—4a5– 6 ( )(h) —32ba–45– 3

2. Simplify each of the following. ( )(c) —4623– 3 ÷ —6432– (d) —((–(—–4)4—6))62—××—((––—55)22—)3
(h) —((bb—22dd34—))23
— — 3 ×–2 4—2 (b) —33—×64—(62—)3 (g) —((mm—52nn—37))2–3
( ) (a) 2


(e) —x2—yx6y—×2 —x3 (f) —(h(h—3kk2)—2)4

3. Simplify each of the following.

(a) — (2m—2n —4 )m—3 7×n— ( 3— m—n4)–2 ( b) (— 5x—y4 —) 2 x—×4y6—6x — y (c) 2—(4dd—53ee—65) ×—× ((—36dd—e3e2–)43)–2

13

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

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

Division in Law of indices Solution Conclusion
index form 23 – 3 = 20 Repeated multiplication from the
solution
(a) 23 ÷ 23 —22 ×—× 22—××–22–
20

(d) m5 ÷ m5 m5 – 5 = m0 ²mm²×× ²mm ²×× mm²×ײmm²××±mm±± m0

(c) 54 ÷ 54

(d) (–7)2 ÷ (–7)2

(e) n6 ÷ n6

Discussion:
1. Are your answers similar to the answers of the other groups?
2. What is your conclusion regarding zero index?

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

20
m0

7KHUHIRUH D QXPEHU RU DQ DOJHEUDLF WHUP ZLWK D ]HUR LQGH[ ZLOO JLYH D YDOXH RI
In general, a0 a 

How do you verify a–n = –a1–n– ?

Brainstorming 5 In groups

A i m : To veri fy a–n = — a n.
Steps:

1. Study and complete the following table.

14

Chapter 1 Indices

Division in Solution Conclusion
index form from the
Law of indices Repeated multiplication solution CHAPTER 1
(a) 23 ÷ 25
23 – 5 = 2–2 —2 — × 2 — ×î— 2 —×î 2 – –× – –2 = – 2 – × – –2 = – 2 –2 2 – 2 = 2– –2
(b) m2 ÷ m5 m2 – 5 = m–3 –m––×—m—×m—m×—×m—m —× m — = —m —× m – – × – – m – = – m –3
(c) 32 ÷ 36 m – 3 = – m – 3–

(d) (– 4)3 ÷ (– 4)7

(e) p4 ÷ p8

Discussion Scan the QR Code or visit
1. Are your answers similar to the anwers of the other groups? http://bukutekskssm.my/
2. What is your conclusion? Mathematics/F3/Chapter1
AlternativeMethod.mp4
)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW to watch a video that
describes alternative
2–2 = 2 — 2 method to verify a± = —a n.
m –3 = m— 3

In general, a–n = –a –n ; a  0 BULLETIN

Example 11 Negative index is a
number or an algebraic
term that has an index of
a negative value.

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

( a) a – 2 (b) x – 4 ( c) – 8– –5– TIPS

(d) –y ––9 – ( e) 2 m –3 ( f) — 5 n –8 Ƈ a–n ±a1±n
Ƈ an ±a1±–±n

( )(g) –2– ± ( )(h) –x– –7 ( ) ( )Ƈ ±ab± –n ±ab± n
3 y
=

2. State each of the following in negative index form. REMINDER
2a –n  2²1an±
(a) — 3 4 ( b) —m 5 (c) 75
SMART MIND
(d) n20 ( )(e) –45– 8 ( )(f) –mn–
( ) ± ² 4 ± = x y
3. Simplify each of the following.
What are the values of x
(a) 32 × 34 ÷ 38 (b) (—24—)2 —× (—35—)3 (c) —(4x—y2—)2 —× x—5y and y?
(28 × 36)2 (2x3y)5
15

CHAPTER 1 Solution: (b) x – 4 = –x1–4 (c) –8–1––5 = 85 (d) –y–1––9 = y9
1. (a) a–2 = a–12– (f ) —35 n– 8 = —53n–8

(e) 2m–3 = m—23 –2– –10 –3– 10 –x– –7 –y– 7

= =
( ) ( )(g) ( ) ( )(h)
32 yx

2. (a) 3—14 = 3– 4 (b) —m15 = m–5 (c) 75 = 7—1–5 (d) n20 = n—–12–0

( ) ( )(e)–4–8 –5– –8 ( ) ( )(f)–m–15 –n– –15

= =
54 nm

(b) —(2(4—2)28—×× (—3365)—)23 (c) —(4x—(y22x—)32y—×)5x—5y TIPS
= —2218—6××—331–512
3. (a) 32 × 34 ÷ 38 = 28 – 16 × 315 – 12 = —42x—2y—4 ×—x5—y1 y0 = 1
= 2–8 × 33 25x15y5 y1 = y
= 32 + 4 – 8 = —2383
= 3–2 = 1—6 x2 + 5 – 15 y4 + 1 – 5
= —1 32

32 = —1 x–8 y0
2

= —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) —2 m–5 ( )(j) – —3 m– 4
7 8
( )(k) —2 –12 ( )(l) – —3 –14 ( )(m) —x –10
5 7 y ( )(n) —2x– – 4 ( )(o) —1 –5
3y 2x

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

(a) —1 (b) —1 (c) —1 (d) —1 (e) 102
54 83 m7 n9
( )(j) —yx 10
(f) (– 4)3 (g) m12 (h) n16 ( )(i) —47 9

3. Simplify each of the following.

(a) (—42(—)436—×)24–5 (b) —((22—3××—3342–))5–3 (c) —(23—)(–52—2×)5–(5–4–)2
(f) (—2m—–2—(n4)m—5 2×—n(43)—2m–4n–)–2
(d) —3m—2n9—4m×—3n(m5—n—3)––2 (e) —(2m—2n—2()9—–m3 —×3n()—32m–n–2–)–4

16

Chapter 1 Indices

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

Relationship between n¥a and a—n1 Determine and state the CHAPTER 1
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: Square roots are used (b) x3 = 64 Ƈ 9 = 32 Ƈ 64 = 43
(a) x2 = 9 to eliminate squares. 3¥x3 = 3¥ 3
x =4 Cube roots are used to
¥x2 ¥ 2 eliminate cubes.
x =3

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
of 2 is —1 .
x2(—21 )= 9—21 x3(—31 )= 64(—31 ) 1 is the reciprocal of a.
x1 = 32(—21 ) 2 x1 = 43(—13 ) —a

x =3 The reciprocal x =4
of 3 is —13 .

From the two methods to determine the values of x in the examples SMART MIND
above, it is found that:
What is the solution for
2¥x = x–12 ¥– 4 ? Discuss.
3¥x = x–13

In general, n¥a = a–1n ; D  0

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

(a) 2¥ E 3¥± F 5¥m (d) 7¥n

2. Convert each of the following terms into the form n¥a .
(a) 125—51 (b) 256—81 (c) (–1 000)—31 (d) n1—12

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

(a) 5¥± E 6¥ F — 13 G ± —51

Solution: (b) 3¥± ± — 31 F 5¥m = m—1 (d) 7¥n = n—1
(b) 25681– = 8¥256 5 7
1. (a) 2¥36 = 36—12
2. (a) 125—15 = 5¥125 (c) (–1 000)—31 = 3¥(–1 000) (d) n—1 = 12¥n
12

17

CHAPTER 1 3. (a) 5¥–32 = (–32)—15 (b) 6¥729 = 729—16 (c) 512 13– = 83(3–1) (d) (–243)—15 = (–3)5(—15 )

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

= –2 =3 <RX FDQ XVH D VFLHQWL¿F
calculator to check the
MIND TEST 1.2g answers.

1. Convert each of the following terms into the form a–1n . (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)—31

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—n1

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

(a) 64—32 (=64420)—9136(—13 ) (64—31 )2 3¥642 (3¥64)2
= 163(—31 ) = 43(—31 )(2) = 3¥4 096 = 42
= 42 = 16
= 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)—n1 = (a—1n )m
a—mn = n¥am = (n¥a)m

18

Chapter 1 Indices

Example 13

1. Convert each of the following into the form (am)–n1 and (a–n1)m. CHAPTER 1

(a) 81—32 (b) 27—23 (c) h—35

2. Convert each of the following into the form n¥am and (n¥a)m.

(a) 343—32 (b) 4 096—56 (c) m—25

Solution: (b) 27—32 = (272)—13 (c) h—35 = (h3)—51
1. (a) 81—32 = (813)—21 27—23 = (27—13 )2 h—53 = (h—15 )3

81—23 = (81—21 )3

2. (a) 343—23 = 3¥ 2 (b) 4 096—56 = 6¥ 5 (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.

—8116 —34 —2
a—mn 729—65 121—32 w—37 ( ) ( )x—25 —hk
3

(am)—1n

(a—n1 )m

n¥ am

(n¥ a )m

Example 14

1. Calculate the value of each of the following.

(a) 9—25 (b) 16—45

Solution: (b) 16—45 16—45 = (4¥16)5 = 25 = 32
1. (a) 9—25 16—45 = 4¥165 = 4¥1 048 576 = 32
Method 1
Method 1 9—25 = (¥9)5 = (3)5 = 243 Method 2
Method 2 9—52 = ¥95 = ¥59 049 = 243

19

MIND TEST 1.2i

CHAPTER 1 1. Calculate the value of each of the following.

(a) 27—23 (b) 32—25 (c) 128—27 (d) 256—38

(e) 64—34 (f) 1 024—25 (g) 1 296—43 (h) 49—23
(l) 10 000—34
(i) 2 401—41 (j) 121—23 (k) 2 197—32

2. Complete the following diagrams with correct values.

(a)

Example 16 Chapter 1 Indices

1. Calculate the value of each of the following. (c) (—24—3—54—×—5——32 )–2 CHAPTER 1
4¥ × ¥ 4
(a) ——49——21 ×—1—25—– —13—– (b) —16——43 —× 8—1–——14–
4¥ î 5¥ (26 × 34)—21 (c) (—24—3—54—×—5—32—)2
4¥ × ¥ 4
Solution: (b) —16——43 —× 8—1–——41– = —284—13——4154—(2×—) ×2–55–—24—–23–(–2)–
(26 × 34)—12 = —343(——415()——58×) ×—525–(—342–)
(a) ——49——21 ×—1—25—– ——13 = —3318—×× 5—534
4¥ î 5¥ = —224—6(3—(—4321—))××—3344—2(–(—21—–14) –) = 38 – 1 × 53 – 4
= —(77—24(—)21——14) ××—5(—53(5–)———3115 )– = —23—×—3––1 = 37 × 5–1
= —357
= —7711—××—55–11– 23 × 32 = —2 1—587–
= 71–1 × 5–1 –1 = 23 – 3 × 3–1 – 2 = 437 —25
= 70 × 5–2
= 1 × —512 = 20 × 3–3
= —1 = 1 × —1

25 33
= —1

27

MIND TEST 1.2j

1. Simplify each of the following. (b) —(m—n(2m)—36×n—3()¥——23m—n)–4 (c) —¥ — x3—yz—2 ×—4—x2z–
(a) —3¥—c2d(—c3–e—3d×—2ce—)13—2d 2—e—32– ¥ x5yz8

2. Calculate the value of each of the following. (c) ——(2—6 ×—3—4 ×—52—)—23——
4¥ î ¥ î 3¥
(a) —¥ —– 4—× 1—14 (b) —(5—–3 ×—3—6)——13 ×—4¥— ––
49 × 121 (125 × 729 × 64)– —13 (f) —64——32 ×—3¥—1—25—× —(2—× ——51 )––3
42 × 4¥
(d) —9—¥5—12—×—3¥—34—3 ×—¥–1–2–1––– (e) —(24—×—36—)—21—× —3¥ — î– —¥ –
(64)—13 × (81)—34 × (14 641)—14 16—43 × 27—31

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

21

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

Example 17 FLASHBACK
&DOFXODWH WKH YDOXH RI ¥ î —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
Calculate the value of Convert each base ¥ î —32 ÷ 6
numbers in index form into prime factors and = 3—12 × (2 × 2 × 3)—23 ÷ (2 × 3)
with different bases. calculate the value by = 3—12 × 2—23 × 2—23 × 3—23 ÷ (21 × 31)
applying laws of indices. = 3—12 + —32 – 1 × 2—32 + —32 – 1
Making a conclusion
¥ î —32 ÷ 6 = 12 = 31 × 22

= 12

Example 18 REMINDER
Calculate the value of x for the equation 3x × 9x + 5 ÷ 34 = 1.
Ƈ ,I am = an
Understanding the Planning a strategy then, m = n
problem
The question is an equation. Ƈ ,I 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. 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 If 3x × 9x + 5 ÷ 34 = 1, = 3–2 × 93 ÷ 34
3x × 32(x + 5) ÷ 34 = 30 then, x = –2
3x + 6 = 0 = 3–2 × 32(3) ÷ 34
3x + 2(x + 5) – 4 = 30 3x = –6
3x + 2x + 10 – 4 = 30 x = —–36– = 3–2 + 6 – 4
x = –2
33x + 6 = 30 = 30 The same value
am = an =1 as the value on
m=n the right side
of the equation.

22

Chapter 1 Indices

Example 19 Checking Answers
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
= 39 × 36

(x – 3)(x + 5) = 0 method. = 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
25m × 5n = 58 and 2m × —21n = 2 variables can be solved
using substitution
Solution: 2m × —21n = 2 method or elimination
25m × 5n = 58 method.
52(m) × 5n = 58
Checking Answers

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. Left Right
From 1 :
2m + n = 8 Left: Right:

n = 8 – 2m 3 25m × 5n 58
= 52(m) × 5n
= 52(3) × 52

= 56 + 2

Substitute 3 into 2 Substitute m = 3 into 1 = 58 The same value
m–n=1
2m + n = 8 2m × —1 = 2
m – (8 – 2m) = 1 2(3) + n = 8 2n
m – 8 + 2m = 1 You can also
m + 2m = 1 + 8 6+n=8 substitute m = 3 Left Right
3m = 9 into equation
m = —9 Left: Right:
3
m=3 n = 8 – 6 2 or 3 . 2m × —21n 2

n=2 = 23 × —212

Hence, m = 3 and n = 2. = 23 × 2–2
= 23 + (–2)
= 21

= 2 The same value

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. Calculate the values of x and y which satisfy the experiment 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 + 2x = 3y
1 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 2 + x = 2(3)
by 2 to equate the x =6–2
x =4
3 FRHI¿FLHQWV RI YDULDEOH x.
Hence, x = 4, y = 3
2 : 1 + 2x = 3y

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
(d) (2x3)5 = 2x15
(g) 32—25 = (2¥32)5 (e) m0n0 = 1 (f) 2a– 4 = —21a–4
(i) (5m—14 )– 4 = 6—m25–
( ) ( )(h)—mn –4 —mn 4

=

24

Chapter 1 Indices

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

5Ƒ × 55 53(Ƒ) CHAPTER 1

( )—51Ƒ 3 512 ÷ 5Ƒ

5—1Ƒ 59 (¥25)Ƒ

—56 —5×25–Ƒ– ( )—1 Ƒ
(5Ƒ)—32 5
(Ƒ¥125)Ƒ

3. Copy and complete the following diagram.

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

Value

Skills Enhancement

1. Simplify each of the following.

(a) (mn4)3 ÷ m4n5 (b) 3x × —16 y4 × (xy)3 F ¥xy × 3¥xy2 × 6¥xy5

2. Calculate the value of each of the following. (c) (256)—83 × 2–3
(f) (125)—32 × (25)– —32 ÷ (625)– —41
(a) 64—13 × 5–3 (b) 7–1 × 125—23
(c) axa8 = 1
(d) 24 × 16– —43 (e) ¥49 × 3–2 ÷ (¥81)–1 (f) 2x = —12610–x
(i) 25x ÷ 125 = —51x
3. Calculate the value of x for each of the following equations.

(a) 26 ÷ 2x = 8 (b) 3– 4 × 81 = 3x

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

(g) 36 ÷ 3x = 81(x – 1) (h) (m2)x × m(x + 1) = m–2

25

Self Mastery

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

(a) 4—31 × 50—23 × 10—53 (b) 5—52 × 20—32 ÷ 10–2 (c) 60—12 × 125—23 ÷ ¥15

2. Calculate the value of x for each of the following equations.

(a) 64x—12 = 27x– —52 (b) 3x—32 = —247 x– —34 (c) 25x– —23 – —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
WKDW WKH WHPSHUDWXUH RI D PHWDO URVH IURP Û& WR TÛ&
according to equation T = 25(1.2)m when the metal

was heated for m seconds. Calculate the difference in
WHPSHUDWXUH EHWZHHQ WKH ¿IWK VHFRQG DQG WKH VL[WK
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 dealers, the price RM55 000

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

( )RM55000—8 n In this situation, n is the number of years
9
.

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

Azmi’s car? State your answer correct to the nearest RM.

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.

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 biggest possible square.

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.
my/Mathematics/F3/
in Sheet A? Chapter1SheetA.pdf
8. Discuss the patterns you identify. to download Sheet A.

Sheet A

Number of axis Index form Number of Index form
of symmetry square
– 20
0 21 1 22
2 4
8 16

27

CHAPTER 1 an Index CONCEPT MAP 54 = 5 × 5 × 5 × 5
m × m × m × m × m = m5
Base 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
36 ÷ 34 = 36 – 4 (34)2= 38 (3a4)3 = 27a12
23 × 25 = 23 + 5

Fractional index Negative index Zero index
a0 = 1 ; a  0
a—n1 = n¥a 8–13 = 3¥8 a–n = —a1n ; a  0 20 = 1
a—mn = (am)—1n = (a—n1 )m 8–32 = (82)–31 = (8–13)2 5–3 = —513 m0 = 1
8–23 = 3¥ 2 = (3¥8)2
a—mn = n¥am = (n¥a)m

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 

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

8. Perform operations involving laws of indices.

9. Solve 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
2 21
11 4 22
121

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.

4. Complete Sheet 1(a). Discuss the patterns of answers obtained with your friends.

5. Present your results. TIPS

Activity 2 115 = 161 051
11n Value
110 1 1 5 10 10 5 1
111 11
112 121 1 +1 +1—11
113 1 331
114 11 161051
115
116 121
117
118 1331
119
1110 14641

Sheet 2(a) 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

Sheet 2

Instructions:
1. Carry out the activity in small groups.
2. Construct the Pascal’s Triangle as in Sheet 2.
3. Take note of 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.
6. Are your answers the same as those of other groups?

29

2CHAPTER Standard Form

What will you learn?

2.1 6LJQL¿FDQW )LJXUHV

2.2 6WDQGDUG )RUP

WWhhyy ddoo yyoouu lleeaarrnn tthhiiss cchhaapptteerr??
‡ ,Q VFLHQWLILF ILHOG YHU\ ELJ RU YHU\ VPDOO QXPEHUV

DUH FRPPRQO\ XVHG )RU H[DPSOH LQ DVWURQRP\
the distance between two stars is usually millions
of kilometres while in the study of particles, the
distance between atoms is extremely small.
‡ 1XPEHUV ZULWWHQ LQ VWDQGDUG IRUP DUH ZLGHO\ XVHG
in the field of science, engineering, astronomy and
so on.

Distance in outer space, such as the distance
between two stars in the galaxy, is
measured in light years. One light year is the
distance travelled by light in one year. One light
year is equal to 9 500 000 000 000 km, that
is 9.5 trillion kilometres. Small units such as
nanometre are used for distances closer to zero.
Do you know that 1 nanometre is equal to
0.000 000 001 metre?

30

EExxpplloorriinngg EErraa
The ancient Greeks used a system based on
myriad that is ten thousand. Ten myriads is
equal to one hundred thousand.
Archimedes (287 BC – 212 BC) created a
system of big numbers up to 108 × 1016.

KWWS EXNXWHNVNVVP P\ 0DWKHPDWLFV )
ExploringEraChapter2.pdf

WORD B A N K

‡ HVWLPDWLRQ ‡ DQJJDUDQ
‡ VLJQL¿FDQW ¿JXUH ‡ DQJND EHUHUWL
‡ VWDQGDUG IRUP ‡ EHQWXN SLDZDL
‡ DFFXUDF\ ‡ NHMLWXDQ
‡ VLQJOH QXPEHU ‡ QRPERU WXQJJDO
‡ URXQG R൵ ‡ SHPEXQGDUDQ
‡ DSSUR[LPDWLRQ ‡ SHQJKDPSLUDQ

31

2.1 Significant Figures

CHAPTER 2 :KDW GRHV VLJQL¿FDQW ¿JXUH PHDQ DQG KRZ GR \RX LEARNING
GHWHUPLQH WKH QXPEHU RI VLJQL¿FDQW ¿JXUHV RI D QXPEHU" STANDARD

We use measurement in many situations in our daily life. Examples of Explain the meaning of
frequently used measurements are length, distance, mass, temperature, VLJQL¿FDQW ¿JXUH DQG
area and speed. KHQFH GHWHUPLQH WKH
QXPEHU RI VLJQL¿FDQW
¿JXUHV RI D QXPEHU

The HVWLPDWLRQ of a measurement can be done using

NP DSSUR[LPDWLRQ )RU H[DPSOH WKH GLVWDQFH EHWZHHQ WKH

(DUWK DQG WKH 0RRQ LV NP 7KLV YDOXH LV DQ

Moon estimation calculated using certain methods and stated as
Earth an approximation.

The GHJUHH RI DSSUR[LPDWLRQ of a measurement to the DFWXDO YDOXH shows the level of accuracy
of the measurement. The skill in making estimations and approximations can help you in many

situations in daily life.

Brainstorming 1 In pairs

$LP Determine the importance of making estimations and approximations in daily life.

6WHSV 50%DISCOUNT

1. Read and understand the situations below.

6LWXDWLRQ 1
Hashim is interested in a shirt sold in a supermarket with a 50% discount.
The original price of the shirt is RM47.90. Hashim estimates the price
of the shirt after discount and takes it to the cashier. The cashier
informs him that the price of the shirt is RM28.70. Hashim argues
that his HVWLPDWLRQ RI WKH SULFH LV QRW PRUH WKDQ 50 ,V Hashim’s
estimation correct?

6LWXDWLRQ 2
0UV 7DQ ZDQWV WR EX\ PHWUHV RI FORWK FRVWLQJ 50 SHU PHWUH WR PDNH FXUWDLQV

She PDNHV DQ HVWLPDWLRQ RI WKH WRWDO SULFH RI WKH FORWK DQG DOORFDWHV 50 ,V WKH
money allocated by Mrs Tan VX൶FLHQW?

'LVFXVVLRQ
1. ,Q WKH WZR VLWXDWLRQV DERYH KRZ GLG +DVKLP DQG 0UV 7DQ PDNH HVWLPDWLRQV RI WKH WRWDO

price?
2. Discuss with your friend the importance of making estimations and approximations.
3. State two other situations that require you to make estimations and approximations.

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW:

$SSUR[LPDWLQJ D YDOXH WR D FHUWDLQ VLJQL¿FDQW ¿JXUH DOORZV XV WR PDNH DQ DFFXUDWH
estimation.

32

Chapter 2 Standard Form

You have understood the importance of making estimation for the purpose of obtaining a value
WKDW LV QHDU WKH H[DFW YDOXH 6LJQL¿FDQW ¿JXUHV DUH XVHG WR REWDLQ WKH DSSUR[LPDWH YDOXH.

7KH VLJQL¿FDQW ¿JXUHV RI DQ LQWHJHU RU GHFLPDO UHIHU WR WKH GLJLWV LQ WKH QXPEHU VWDWHG
DFFXUDWHO\ WR D FHUWDLQ GHJUHH RI DFFXUDF\ DV UHTXLUHG 7KH QXPEHU RI VLJQL¿FDQW QXPEHUV LV
counted starting from a non-zero digit.

Brainstorming 2 In pairs CHAPTER 2

$LP 'HWHUPLQH WKH H൵HFW RI WKH SRVLWLRQ RI WKH ]HUR GLJLW LQ LQWHJHUV DQG GHFLPDOV

6WHSV
1. Study the integer cards below.



&DUG &DUG &DUG &DUG

'RHV WKH SRVLWLRQ RI WKH ]HUR GLJLW KDYH DQ\ H൵HFW RQ WKH YDOXH RI GLJLW "

2. Study the decimal cards below.



&DUG &DUG &DUG &DUG

'RHV WKH SRVLWLRQ RI WKH ]HUR GLJLW KDYH DQ\ H൵HFW RQ WKH YDOXH RI GLJLW "

3. Study the decimal cards below.



&DUG &DUG &DUG &DUG

'RHV WKH SRVLWLRQ RI WKH ]HUR GLJLW KDYH DQ\ H൵HFW RQ WKH YDOXH RI GLJLW "

'LVFXVV ZLWK \RXU IULHQG WKH H൵HFW RI WKH SRVLWLRQ RI WKH ]HUR GLJLW RQ WKH YDOXH RI GLJLW

LQ &DUG WR &DUG DQG WKH H൵HFW RI DGGLQJ ]HUR GLJLWV RQ WKH YDOXH RI GLJLW LQ &DUG WR
Card 12.

5. Present the results of your discussion. Compare your results with other pairs.

'LVFXVVLRQ
What is your conclusion concerning the position of the zero digit in an integer or decimal?

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW: FLASHBACK

(a) &DUG &DUG &DUG &DUG &DUG DQG &DUG )RU GLJLW LQ WKH QXPEHU
Ɣ The position of the zero digit between or at the end of the ;
Ƈ 3ODFH YDOXH ± KXQGUHG
QXPEHU PDLQWDLQV WKH SODFH YDOXH RI GLJLW . Ƈ 'LJLW YDOXH ±
(b) Card 4 and Card 8

Ɣ 7KH SRVLWLRQ RI WKH ]HUR GLJLW DV WKH ¿UVW GLJLW KDV FKDQJHG WKH
SODFH YDOXH RI GLJLW .
(c) Card 9, Card 10, Card 11 and Card 12

Ɣ The position of the zero digit at the end of the decimal does not
change the place value of digit 2.

33

CHAPTER 2 ,Q JHQHUDO, TIPS

Ɣ $OO QRQ ]HUR GLJLWV DUH VLJQL¿FDQW ¿JXUHV. Ƈ =HURV EHWZHHQ
Ɣ 7KH GLJLW ]HUR EHWZHHQ QRQ ]HUR GLJLWV LV D VLJQL¿FDQW ¿JXUH. QRQ ]HUR GLJLW DUH
Ɣ 7KH GLJLW ]HUR DW WKH HQG RI DQ LQWHJHU LV D VLJQL¿FDQW ¿JXUH VLJQL¿FDQW ¿JXUHV
)RU H[DPSOH
according to the level of accuracy required.
D
Ɣ 7KH GLJLW ]HUR DW WKH HQG RI D GHFLPDO LV D VLJQL¿FDQW ¿JXUH EHFDXVH (5 VLJQL¿FDQW ¿JXUHV
it determines the level of accuracy of the decimal.
E
Ɣ 7KH GLJLW ]HUR EHIRUH WKH ¿UVW QRQ ]HUR GLJLW LV QRW D VLJQL¿FDQW (6 VLJQL¿FDQW ¿JXUHV
¿JXUH.

+RZ GR \RX GHWHUPLQH WKH QXPEHU RI VLJQL¿FDQW ¿JXUHV" Ƈ )RU D GHFLPDO DOO GLJLWV
EHIRUH QRQ ]HUR GLJLW DUH
'HFLPDO QRW VLJQL¿FDQW ¿JXUHV
)RU H[DPSOH
1RW VLJQL¿FDQW ¿JXUH 6LJQL¿FDQW ¿JXUH
Used only to determine The digit zero between D
or at end of decimal is a (1 VLJQL¿FDQW ¿JXUH
place value of digit 5.
VLJQL¿FDQW ¿JXUH E
VLJQL¿FDQW ¿JXUHV
0 . 0 0 5 014 0 0

6LJQL¿FDQW ¿JXUH Ƈ )RU D ZKROH QXPEHU
$OO QRQ ]HUR GLJLWV DUH VLJQL¿FDQW ¿JXUHV ]HUR DW WKH HQG RI
WKH QXPEHU LV QRW D
,QWHJHU 803 000 VLJQL¿FDQW ¿JXUH XQOHVV
VWDWHG RWKHUZLVH
6LJQL¿FDQW 6LJQL¿FDQW ¿JXUH depending )RU H[DPSOH
¿JXUH on level of accuracy required.
D
Example 1 (5RXQGHG Rႇ WR

'HWHUPLQH WKH QXPEHU RI VLJQL¿FDQW ¿JXUHV IRU WKH QXPEHUV EHORZ VLJQL¿FDQW ¿JXUHV
E
D E F G
(5RXQGHG Rႇ WR
VLJQL¿FDQW ¿JXUH

H I J K

Solution:

D [4 s.f.]

(b) 5 008 [4 s.f.] 7KH GLJLW ]HUR EHWZHHQ QRQ ]HUR GLJLW LV D VLJQL¿FDQW ¿JXUH
7KH GLJLW ]HUR EHWZHHQ QRQ ]HUR GLJLW LV D VLJQL¿FDQW ¿JXUH
(c) 7 409 [4 s.f.] ,I OHYHO RI DFFXUDF\ LV WR WKH QHDUHVW WKRXVDQG
,I OHYHO RI DFFXUDF\ LV WR WKH QHDUHVW KXQGUHG
(d) (i) 15 000 [2 s.f.] ,I OHYHO RI DFFXUDF\ LV WR WKH QHDUHVW WHQ
,I OHYHO RI DFFXUDF\ LV WR WKH QHDUHVW RQH
(ii) 15 000 > V I @
(iii) 15 000 [4 s.f.] 7KH GLJLW ]HUR EHIRUH ¿UVW QRQ ]HUR GLJLW LV QRW VLJQL¿FDQW
¿JXUH
(iv) 15 000 [5 s.f.]
All zeros after non-zero digit at end of decimal are
H [4 s.f.] VLJQL¿FDQW ¿JXUHV
(f) 0.0809 > V I @
(g) 12.051 [5 s.f.]

(h) 1.2700 [5 s.f.]

MIND TEST 2.1a TIPS

1. 6WDWH WKH QXPEHU RI VLJQL¿FDQW ¿JXUHV IRU WKH IROORZLQJ QXPEHUV 6LJQL¿FDQW ¿JXUH FDQ EH
ZULWWHQ DV V I
D E F
G
H I J K

34

Chapter 2 Standard Form

+RZ GR \RX URXQG R൵ D QXPEHU WR FHUWDLQ QXPEHUV RI LEARNING
VLJQL¿FDQW ¿JXUHV" STANDARD

'R \RX VWLOO UHPHPEHU KRZ WR URXQG R൵ D QXPEHU WR D FHUWDLQ SODFH 5RXQG Rႇ D QXPEHU WR
YDOXH" 7KH VDPH FRQFHSW DQG PHWKRG DUH XVHG WR URXQG R൵ D QXPEHU WR FHUWDLQ QXPEHUV RI
VLJQL¿FDQW ¿JXUHV
D FHUWDLQ QXPEHU RI VLJQL¿FDQW ¿JXUHV

Example 2 CHAPTER 2

5RXQG R൵ HDFK RI WKH IROORZLQJ QXPEHUV WR VLJQL¿FDQW ¿JXUHV TIPS

D E F )RU LQWHJHUV WKH GHFLPDO

Solution: SRLQW LV SODFHG EHKLQG
WKH ODVW GLJLW
(a)
WKXV GLJLW UHPDLQV XQFKDQJHG
12
4, 7 and 9 are placed before decimal point. Thus, replace 4, 7 and

}}
'LJLW WR EH URXQGHG R൵ 9 with zero.
}}
7KXV V I FLASHBACK

(b) 7 > 5, thus add 1 to 4. 5RXQG Rႇ WR WKH
(a) QHDUHVW KXQGUHG
12 7 and 6 are placed before decimal point. E QHDUHVW WKRXVDQG
Solution:
2 476 D

Thus, replace 7 and 6 with zero. !
'LJLW WR EH URXQGHG R൵

7KXV V I E (U H PDLQ XQFKDQJHG)

(c) WKXV DGG WR

12 TIPS

)RU LQWHJHUV WKH ¿UVW
} QRQ ]HUR GLJLW LV D
DQG DUH SODFHG EHIRUH GHFLPDO VLJQL¿FDQW ¿JXUH
}

'LJLW WR EH URXQGHG R൵ SRLQW 7KXV UHSODFH DQG ZLWK ]HUR

7KXV V I

Example 3 E VLJQL¿FDQW ¿JXUH QU I Z
9 > 5, thus add 1 to 7.
5RXQG R൵ WR :K\ VKRXOG WKH GLJLWV
D VLJQL¿FDQW ¿JXUHV DIWHU WKH GLJLW WKDW LV
Solution:
(a) URXQG HG Rႇ LQ D GHFLPDO

EH GURSSHG"

1 2 7 9
}
6 8.

Digit 9 is placed after decimal point. Thus, 9 is dropped.

'LJLW WR EH URXQGHG R൵

7KXV V I

35

(b) 8 > 5, thus add 1 to 6.

1 Digit 8 is placed before decimal point. Thus, 8 is replaced
with zero. 7 and 9 are dropped.
6 8. 79

'LJLW WR EH URXQGHG R൵
}}
7KXV V I
}
CHAPTER 2 Example 4 E VLJQL¿FDQW ¿JXUHV QU I Z
5RXQG R൵ WR
D VLJQL¿FDQW ¿JXUHV 5RXQG Rႇ WR
Solution: VLJQL¿FDQW ¿JXUH DQG
(a) 1 2 VLJQL¿FDQW ¿JXUHV

0.00 8 0 2 5 WKXV DGG WR

'LJLW WR EH URXQGHG R൵ Digit 5 is dropped because it is placed after the decimal
point.

7KXV V I

(b) 2 < 5, thus digit 0 remains unchanged.

12 Digits 2 and 5 are dropped because it is placed after the
decimal point.
0.00 8 0 2 5
'LJLW WR EH URXQGHG R൵

7KXV V I

MIND TEST 2.1b

1. &RPSOHWH WKH WDEOH EHORZ E\ URXQGLQJ R൵ HDFK QXPEHU EHORZ WR WKH JLYHQ VLJQL¿FDQW ¿JXUH.

1XPEHU 3 VLJQL¿FDQW ¿JXUHV 2 VLJQL¿FDQW ¿JXUHV 1 VLJQL¿FDQW ¿JXUH

D
(b) 5 261

F
(d) 20.68
(e) 8.595
(f) 5.9

J
(h) 0.09184
(i) 0.005709

2. Calculate each operation below. 6WDWH WKH DQVZHU WR WKH VLJQL¿FDQW ¿JXUHV VKRZQ LQ WKH
brackets.

(a) 2.57 × 4.5 + 0.45 [4] (b) 8.59 ÷ 2.1 – 1.26 > @

F ± î > @ G · [2]

H î · > @ (f) 10.25 ÷ 0.75 – 4.2 × 0.2 [2]

J ± î [1] (h) 4.94 + 5 .76 ÷ 0.26 × 1.4 > @

36

Chapter 2 Standard Form

2.2 Standard Form

+RZ GR \RX UHFRJQLVH DQG ZULWH QXPEHUV LQ VWDQGDUG LEARNING
IRUP" STANDARD

0DQ\ VFLHQWL¿F ¿HOGV VXFK DV DVWURQRP\ ELRORJ\ SK\VLFV DQG 5HFRJQLVH DQG ZULWH
engineering frequently use numbers that are too big or too small in their QXPEHUV LQ VWDQGDUG IRUP

research. These numbers are written in standard form to make writing CHAPTER 2

easier.

6WDQGDUG IRUP is a way to write a VLQJOH QXPEHU in the form;
A × 10Q

where ” A < 10 and Q is an integer.
)RU H[DPSOH WKH ODQG DUHD RI 0DOD\VLD LV P2. This value can be written as
î 11 m2 RU î 11 m2 RU GHSHQGLQJ RQ WKH QXPEHU RI VLJQL¿FDQW ¿JXUHV UHTXLUHG

+RZ GR \RX FKDQJH D VLQJOH QXPEHU WR VWDQGDUG IRUP" FLASHBACK

When a single number is changed to standard form: Ƈ DQ LV D SRVLWLYH LQGH[
Ƈ D –Q LV D QHJDWLYH LQGH[
‡ 1XPEHUV ZLWK YDOXH PRUH WKDQ LV ZULWWHQ DV D SRVLWLYH LQGH[.
‡ 1XPEHUV ZLWK YDOXH OHVV WKDQ LV ZULWWHQ DV D QHJDWLYH LQGH[.

Example 5

Write the following single numbers in standard form.

D E F

Solution: E î 100 F î 1 000
(a) î 10 î 102 î 10

Place value Decimal point after Place value is hundreds Place value is thousands
is tens ¿UVW QRQ ]HUR GLJLW.

Example 6 FLASHBACK

Write the following decimals in standard form. ² D1Q± D–Q

D E F G DISCUSSION CORNER

Solution: Is × 100 D QXPEHU
LQ VWDQGDUG IRUP"
D î —110 E î —1 0–10—0 'LVFXVV
î 10–1 î —110–

Place value is one tenths î 10±

Place value is one thousandths

37

F î —101–0 G î —1 0–10—0
î —110–2 î —110–

î 10–2 î 10±

Place value is one hundredths Place value is one thousandths

CHAPTER 2 +RZ GR \RX FKDQJH D QXPEHU LQ VWDQGDUG IRUP WR VLQJOH QXPEHU"

When a number in standard form is changed to a single number:
‡ 7KH QXPEHU ZLOO EH HTXDO WR RU PRUH LI WKH LQGH[ LV SRVLWLYH
‡ 7KH QXPEHU ZLOO EH OHVV WKDQ LI WKH LQGH[ LV QHJDWLYH

Example 7 FLASHBACK

Write 4.17 × 105 as a single number. 105 [ [ [ [
10± 1±10±5±
Solution:
4.17 × 105 î


Example 8 BULLETIN

Write î í as a single number. WHUD
QDQR
Solution:
î í î ²10²010²00±


Example 9 SMART MIND

Determine terabytes in bytes. State the answer in standard form. :KDW LV WKH YDOXH RI
Solution: WHUD LQ QDQR"

terabytes î 12 bytes
î ) × 1012 bytes
Use index law DP × DQ DP + Q
î ) bytes
î 15 bytes

Example 10

Determine 0.0057 nanometre in metre. State your answer in standard form.

Solution: Use index law DP × DQ DP + Q

0.0057 nanometre î í metre
î í ) × 10í metre
î í í ) metre
î í í ) metre
î í metre

38

Chapter 2 Standard Form

Brainstorming 3 In pairs

$LP Write metric measurements in standard form.
Steps

1. Complete the table below by writing the single numbers for metric measurements in
standard form.

3UH¿[ 6\PERO 9DOXH 6WDQGDUG IRUP CHAPTER 2
exa 1 × 1018
peta E 6LQJOH QXPEHU
tera P 1 000 000 000 000 000 000 1 × 100
giga T 1 × 10–1
mega G 1 000 000 000 000 000
kilo M 1 000 000 000 000
hecto k 1 000 000 000
deca h 1 000 000
– da 1 000
deci – 100
centi d 10
milli c 1
micro m 0.1
nano P 0.01
pico n 0.001
femto p 0.000 001
atto f 0.000 000 001
a 0.000 000 000 001

0.000 000 000 000 001
0.000 000 000 000 000 001

'LVFXVVLRQ

A number which is too big or too small in value can be written as a single number or in standard
form. Which form will you choose for an arithmetic operation? Give your reasons.

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW:

Standard form makes it easier to write very big and very small numbers in a form that
is simple and easy to understand.

MIND TEST 2.2a TIPS

1. Write the following single numbers in standard form. 8VH GDWD IURP

D E F G %UDLQVWRUPLQJ WR
VROYH TXHVWLRQ
H I J K

2. Change the numbers in standard form to single numbers.

(a) 2.5 × 100 E × 101 F × 102
(e) 9.1 × 104 (f) 6.2 × 10–1
(d) 5.07 × 10 (i) 8.504 × 10– 4

(g) 7.29 × 10–2 K × 10±

3. Change the following metric measurements to the units given in the brackets. State your

answers in standard form.

(a) 1 050 kilometres [metre] (b) 216 gigabytes [byte]

(c) 0.75 teralitre [litre] (d) 95 micrometres [metre]

H nanometres [metre] (f) 0.089 femtometre [metre]

39

+RZ DUH EDVLF DULWKPHWLF RSHUDWLRQV LQYROYLQJ QXPEHUV LEARNING
LQ VWDQGDUG IRUP SHUIRUPHG" STANDARD

2SHUDWLRQV RI DGGLWLRQ DQG VXEWUDFWLRQ 3HUIRUP EDVLF DULWKPHWLF
RSHUDWLRQV LQYROYLQJ
QXPEHUV LQ VWDQGDUG IRUP

Example 11

Calculate the value of each of the following operations. State your answer in standard form.

CHAPTER 2 D î + 5.92 × 10 (b) 4.27 × 105 î 5
(c) 7.02 × 104 + 2.17 × 105 (d) 9.45 × 106 ± î 5

Solution:

D î + 5.92 × 10 (b) 4.27 × 105 î 5 FLASHBACK
î î 5
î î 5 Ƈ 5DQ DQ
î î 5 DQ
)DFWRULVH 10 î 1 × 105 DQ
î 1 + 5 Ƈ î Q î Q
î 6 Q
Q)

(c) Method 2 TIPS

Method 1 7.02 × 104 + 2.17 × 105 )RU RSHUDWLRQV LQYROYLQJ
î 10–1 × 105 + 2.17 × 105 DGGLWLRQ DQG VXEWUDFWLRQ
7.02 × 104 + 2.17 × 105 î 5 + 2.17 × 105 FKDQJH LQGH[ ZLWK VPDOO
î 5 YDOXH WR LQGH[ ZLWK ODUJH
î 4 + 2.17 × 101 × 104 î 5 YDOXH DV LQ PHWKRG
î 4 + 21.7 × 104 RI H[DPSOH F DQG
H[DPSOH G
î 4
SMART MIND
î 4

î 1 × 104

î 1 + 4 10൥ change to 101 × 104
î 5 to simplify calculation.

(d) Method 1 Method 2 &DOFXODWH WKH IROORZLQJ
ZLWKRXW XVLQJ D FDOFXODWRU
9.45 × 106 ± î 5 9.45 × 106 ± î 105 Ƈ î 3 î 5
î 6 ± î 10–1 × 106 Ƈ î ± î
î 6 ± î 6
î 101 × 105 ± î 5 ± î 6
î 5 ± î 5 î 6

± î 5

î 5

î 1 × 105

î 1 + 5

î 6

40