BUKU TEKS ADDMATHS FORM 4 (ANSWERS)

Substitution y = 2 into 1.
3

( )5 2 +x=1 7
3 3
x = –

Substition y = 1 into 1.

5(1) + x = 1

x = −4

Thus, x = − 7 , y = 2 and x = −4, y = 1 are the solutions to these simultaneous equations.
3 3

Kaedah penggantian: …1
5y + x = 1 …2
x + 3y2 = −1

From 1, x = 1 − 5y …3

Substitute 3 into 2.

1 − 5y + 3y2 = −1

3y2 − 5y + 2 = 0

( 3y – 2)(y – 1) = 0

3y – 2 = 0 or y–1=0
2 y=1
y= 3

Substitute y = 2 into 3.
3

x = 1 − 5( 2 )
3
= − 73

Substitute y = 1 into 3.
x = 1 – 5(1)
= −4

Thus, x = − 7 , y = 2 and x = −4, y = 1 are the solutions to these simultaneous equations.
3 3

Graph represented:

Thus, x = − 7 , y = 2 and x = −4, y = 1 are the solutions to these simultaneous equations.
3 3

17

(c) Substitution method:
y = 3 – x …1

1 – 1 =2 …2
x y

Substitution 1 into 2.

1 – 1 = 2 …3
x 3–x

3 × x(3 − x): 3 − x − x = 2x(3 − x)

3 − 2x = 6x − 2x2

2x2 − 8x + 3 = 0
–(–8) ± ! (–8)2 – 4(2)(3)
x= 2(2)

= 8 ± ! 40
4

x= 8 + ! 40 x= 8 – ! 40
4 4

= 3.5811 = 0.4189

Substitution x = 3.5811 into 1.
y = 3 − 3.5811
= −0.5811

Substitution x = 0.4189 into 1.
y = 3 − 0.4189
= 2.5811

Thus, x = 3.5811, y = −0.5811 and x = 0.4189, y = 2.5811 are the solutions to these
simultaneous equations.

Graph represented:

Thus, x = 3.5811, y = −0.5811 and x = 0.4189, y = 2.5811 are the solutions to these
simultaneous equations.

18

(d) Elimination method:
3x + 5y = 1 …1
4
x + 2y = y …2

2 × 3: 3x + 6y = 12 …3
y

3 – 1: y= 12 –1
y

y2 + y – 12 = 0 y–3=0
(y + 4)(y – 3) = 0 y=3

y + 4 = 0
y = –4

Substitution y = –4 into 1.
3x + 5(–4) = 1

3x = 21
x=7

Substitution y = 3 into 1.
3x + 5(3) = 1
3x = –14
x = – 134

 134 , y = 3 are the solutions to these simultaneous equations.

Substitution method: …1
…2
3x + 5y = 1
4
x + 2y = y

x= 4 − 2y …3
y

Substitute 3 into 1.

( )34 + 5y = 1
y – 2y

12 – 6y + 5y = 1
y
12
y –y–1= 0

y2 + y – 12 = 0

(y + 4)(y – 3) = 0

y + 4 = 0 or y – 3 = 0

y = –4 y=3

19

Substitute y = −4 into 3.

x= 4 – 2(–4)
– 4

=7

Substitute y = 3 into 3.

x= 4 – 2(3)
3

= – 134

Thus, x = 7, y = −4 and x = – 14 , y = 3 are the solutions to these simultaneous equations.
3
Graph representation:

Thus, x = 7, y = −4 and x = – 14 , y = 3 are the solutions to these simultaneous equations.
3

(e) Substitution method: …1

2x + 4y = 9
4x2 + 16y2 = 20x + 4y −19 …2

From 1, x= 9 – 4y …3
2

SuSbusbtistutitiuotne 3 into 2.

9 – 4y 9 – 4y
2 2
( ) ( )4
2 + 16y2 = 20 + 4y −19

81 – 72y + 16y2
4
( ) 4 + 16y2 = 10(9 – 4y) + 4y −19

81 – 72y + 16y2 + 16y2 = 90 – 40y + 4y – 19

32y2 – 36y + 10 = 0

16y2 – 18y + 5 = 0

(8y – 5)(2y – 1) = 0

8y – 5 = 0 or 2y – 1 = 0

y= 5 y= 1
8 2

20

Substitute y = 5 into 3.
8

( )x = 9–4 5
2 8

= 13
4

Substitute y = 1 into 3.
2

( )x = 9–4 1
2 2

= 7
2

MThakuas, x = 13 , y= 5 and x= 7 , y= 1 are the solutions for these simultaneous equation.
4 8 2 2

Graph representation:

MThakuas, x = 13 , y= 5 and x= 7 , y= 1 are the solutions for these simultaneous equation.
4 8 2 2

(f) Substitution method:
x + y – 4 = 0 …1

x2 – y2 – 2xy = 2 …2

From 1, x = 4 – y…3

Substitute 3 into 2.

(4 – y)2 – y2 – 2y(4 – y) = 2

16 – 8y + y2 – y2 – 8y + 2y2 = 2

2y2 – 16y + 14 = 0

y2 – 8y + 7 = 0

(y – 1)(y – 7) = 0

y – 1 = 0 y – 7 = 0

y=1 y=7

21

Substitute y = 1 into 3.
x=4–1
=3

Substitute y = 7 into 3.
x=4–7
= –3

Thus, x = 3, y = 1 and x = –3 , y = 7 are the solutions for these simultaneous equation.

Graph representation:

Thus, x = 3, y = 1 and x = –3 , y = 7 are the solutions for these simultaneous equation.

2. (a)

y

30 1 2 3 4 5x
25
20
15
10
5

–5 –4 –3 –2 –1–50
–10
–15
–20
–25

The solutions for these simultaneous equation are (–4.3, –1.7) and (4.0, 2.5).

22

(b) y

6

5

4

3

2

1

–4 –3 –2 –1 0 x
–1 12 34

–2

–3
–4

–5

The solutions to these simultaneous equations are (–1.2, –3.2) and (2.9, 0.9).
Self Practice 3.5 (Page 84)
1. Let x represents the length while y represents the width.

xy = 72 …1
2x + 2y = 34 …2

DariFpraodma 1, x= 72 …3
y

SGuabnsttiitkuatne 3 kinetodalam 2.
( )272
y + 2y = 34…4

4 × y: 144 + 2y2 = 34y

2y2 − 34y + 144 = 0

y2 − 17y + 72 = 0

(y − 9)(y − 8) = 0

y – 9 = 0 or y – 8 = 0y
y=8
=9

Substitute y = 9 into 3.

x= 72
9

=8

Substitute y = 8 into 3.

x= 72
8

=9

Thus, the possible length and width of the plank are 8 cm and 9 cm respectively.

23

2. Perimeter of the pond: 2y + 2(12 − x) = 20
2y + 24 − 2x = 20

−2x + 2y = −4
−x + y = −2…1

10x + (10 − y)(12 − x) = 96

10x + 120 − 10x − 12y + xy = 96

xy − 12y + 120 = 96
xy − 12y = −24…2

From 1, y = x − 2…3

Substitute 3 into 2. x–6=0
x=6
x (x − 2) − 12(x − 2) = −24
x2 − 2x − 12x + 24 = −24
x2 − 14x + 48 = 0

(x − 8)(x − 6) = 0
x – 8 = 0
x=8

Substitute x = 8 into 3.
= 8 −2y
=6

Substitute x = 6 into 3.
= 6 −2y
=4

Thus, the values of x and y are 8 cm and 6 cm or 6 cm and 4 cm respectively.

Intensive Practice 3.2 (Page 84)

1. (a) x − 3y + 4 = 0 …1
x2 + xy − 40 = 0 …2

From 1, x = 3y − 4 …3

Substitute 3 into 2.

(3y − 4)2 + (3y − 4)y − 40 = 0
9y2 − 24y + 16 + 3y2 − 4y − 40 = 0

12y2 − 28y − 24 = 0
3y2 − 7y − 6 = 0

(y − 3)(3y + 2) = 0
 
2
3

24

Substitute y = 3 into 3.
x = 3(3) − 4
=5

Substitute y = – 2 into 3.
3

( )x= 3 –  2 −4
3

= −6 2
3
Thus, x = 5, y = 3 and x = −6, y = – are the solutions to these linear equations.

(b) k − 3p = −1 …1
p + pk − 2k = 0 …2

From 1, k = 3p − 1 …3

Substitute 3 into 2.

p + p(3p − 1) − 2(3p − 1) = 0

p + 3p2 − p − 6p + 2 = 0

3p2 − 6p + 2 = 0
–(–6) ± ! (–6)2 – 4(3)(2)
p= 2(3)

p= 6 + ! 12 atauo r p= 6 – ! 12
6 6

= 1.5774 = 0.4226

Substitute p = 1.5774 into 3.
k = 3(1.5774) − 1
= 3.7322

Substitute p = 0.4226 into 3.
k = 3(0.4226) − 1
= 0.2678

Thus, k = 3.7322, p = 1.5774 and k = 0.2678, p = 0.4226 are the solutions to
these simultaneous equations.

2. x – 2y = 1.…1
y x
2x + y = 3.…2

From 2, y = 3 − 2x…3

Substitute 3 into 1.

x – 2(3 – 2x) = 1…4
– 2x x
3

25

4 × x(3 − 2x): x2 − 2(3 − 2x)2 = x(3 − 2x)

x2 − 2(9 − 12x + 4x2) = 3x − 2x2

x2 − 18 + 24x − 8x2 = 3x − 2x2

−5x2 + 21x − 18 = 0

5x2 − 21x + 18 = 0

(5x − 6)(x − 3) = 0 x – 3 = 0x
5x – 6 = 0 or x=3

= 6
5

Substitute x = 6 into 3.
5
2( 6 )
y = 3 − 5

= 3
5

Substitute x = 3 into 3.

y = 3 − 2(3)

= −3 6 3
( )Thus, the intersection coordinates are 5 5
, and (3, –3).

3. Substitute(–2, 2) into the equation.

2+ 1 (2) = h
2 2

h = −2

1 + 2 = k
(–2) 2
1
k = 2

x + 1 y = −1 …1
2

1 + 2 = 1 …2
x y 2

From 1, y = −2x − 2…3

Substitute 3 into 2.

1 + 2 = 1 …4
x (–2x – 2) 2

4 × x(–2x – 2): –2x – 2 + 2x = 1 x(–2x – 2)
2
–2 = –x2 – x

x2 + x – 2 = 0

(x – 1)(x + 2) = 0

x – 1 = 0 or x + 2 = 0

x=1 x = –2

Substitute x = 1 into 3.
y = −2(1) − 2

= −4

26

Substitute x = −2 into 3.
y = −2(−2) − 2

=2

Thus, x = 1, y = −4 is another solution for these simultaneous equation.

4. x2 + (x + y)2 = (2x + 3)2… 1
x + x + y + 2x + 3 = 30
4x + y = 27 …2

From 2, y = −4x + 27…3

Substitute 3 into 1.
x2 + (x − 4x + 27)2 = (2x + 3)2
x2 + (−3x + 27)2 = (2x + 3)2
x2 + 9x2 − 162x + 729 = 4x2 + 12x + 9

6x2 − 174x + 720 = 0
x2 − 29x + 120 = 0

(x − 24)(x − 5) = 0
x – 24 = 0 or x – 5 = 0 x
= 24 x = 5

Substitute x = 24 into 3.
y = −4(24) + 27

= −69 (Ignore)
Substitute x = 5 into 3.
y = −4(5) + 27

=7

Thus, the possible value for x and y are x = 5, y = 7 respectively.

5. 2x2 + 4xy = 66 …1
8x + 4y = 40
2x + y = 10 …2

From 2, y = −2x + 10…3

Substitute 3 into 1.

2x2 + 4x(−2x + 10) = 66

2x2 − 8x2 + 40x = 66

−6x2 + 40x − 66 = 0

3x2 − 20x + 33 = 0

(3x − 11)(x − 3) = 0

3x – 11 = 0 or x–3=0
x=3
x= 11
3

27

SGubanstitkuatne x = 11 into 3.
3

( )y 11
= −2 3 + 10

= 8
3

Substitute x = 3 into 3
y = −2(3) + 10

=4

IVsiopluamdue = 11 × 11 × 8 aOtaru IVsiopluamdue = 3 × 3 × 4
3 3 3 = 36 cm3

= 35.8519 cm3

MThauksa,,tihsei paodssuibylaenvgomluumnegkfoinr tbhaegciukbuobiodidareitu35ia.8la5h1935c.m853 1o9r 3c6mc3mat3au 36 cm3.
6. 2x2 + 11y2 + 2x + 2y = 0… 1

x − 3y + 1 = 0… 2

From 2, x = 3y − 1…3

Substitute 3 into 1.

2(3y − 1)2 + 11y2 + 2(3y − 1) + 2y = 0

2(9y2 − 6y + 1) + 11y2 + 6y − 2 + 2y = 0

18y2 − 12y + 2 + 11y2 + 6y − 2 + 2y = 0

29y2 − 4y = 0

y(29y − 4) = 0 4
29
y = 0 aotaru y =

Substitute y = 0 into 3.
x = 3(0) − 1

= −1

SGuabnstiktuatne y = 4 into 3.
29

( )x= 3 4 −1
29

= – 1279

Thus, the intersection points between the movement of the fish and the boat are (−1, 0) and
4
17( )–,29 .
29

7. 2x2 + 4y2 + 3x – 5y = 25 …1
y – x + 1 = 0 …2

From 2, y = x – 1…3

28

2x2 + 8x2 – 8x + 4 + 3x – 5x + 5 = 25

6x2 – 10x – 16 = 0

3x2 – 5x – 8 = 0

(3x – 8)(x + 1) = 0
8
x = 3 and x = −1

Substitute x = −1 into 3.

y = −1 – 1

= –2

Substitute x = 8 into 3.
3

y= 8 –1
3

= 5
3

Thus, the intersection points between both sailing boat and speedboat are (–1, −2) and
( ).
8 , 5
3 3

Mastery Practice (Page 86)

1. (a) Let x represents History book, y represents Mathematics books and z represents Science
books.

x + 2y + 3z = 120
2 x + 3y + 2z = 110
x + 4y + 2z = 180

(b) Let x represents 10 sen coin, y represents 20 sen coin and z represents 50 sen coin.



x − 3y – 2z = 25
2. (a) x − y + 2z = 3 …1

x + y − 3z = −10 …2
2 x + y − z = −6 …3

2 − 1: 2y − 5z = −13 …4
1 × 2: 2x − 2y + 4z = 6 …5
3 − 5: 3y − 5z = −12 …6
6 − 4:
y=1

Substitute y = 1 into 4.
2(1) − 5z = −13
−5z = −15
z=3

29

Substitute y = 1 and z = 3 into 1.
x − 1 + 2(3) = 3
x = −2

Thus, x = −2, y = 1 and z = 3 are the solutions to this system of linear equations.

(b) x + 2y + 5z = −17 …1
2 x − 3y + 2z = −16 …2
3x + y − z = 3 …3

1 × 3: 3x + 6y + 15z = −51 …4
2 × 2: 4x – 6y + 4x = –32 …5
4 + 5: 7x + 19z = –83 …6
3 × 3: 9x + 3y – 3z = 9 …7
2 + 7: 11x – z = –7 …8

z = 11x + 7…9

Substitute 9 into 6.

7x + 19(11x + 7) = –83
7x + 209x + 133 = –83

216x = –216
x = –1

Substitute x = –1 into 9.
z = 11(–1) + 7
= –4

Substitute x = –1 and z = –4 into 1.
–1 + 2y + 5(–4) = −17
2y = 4
y=2

Thus, x = −1, y = 2 and z = −4 are the solutions to this system of linear equations.

3. Let x represents the first angle, y represents the second angle and z represents the third angle.
x + y + z = 180 …1
y − 50 = 4x …2
…3
z − 40 = x

From 2, y = 4x + 50…4

From 3, z = x + 40 …5

Substitute 4 and 5 into 1.

x + 4x + 50 + x + 40 = 180
6x + 90 = 180
6x = 90
x = 15

30

Substitute x = 15 into 3.
z – 40 = 15

z = 55

Substitute x = 15 into 2.
y − 50 = 4(15)
y = 110

Thus, x = 15, y = 110 and z = 55 are the solutions for to this system of linear equations.

4. h(x – y) = x + y – 1 …1
x + y – 1 = hx2 – 11y2 …2

Substitute (5, h) into 2.

5 + h – 1 = h(5)2 – 11(h)2

5 + h – 1 = 25h – 11h2

11h2 – 24h + 4 = 0

(11h – 2)(h – 2) = 0

11h – 2 = 0 or h–2=0
2 h=2
h= 11

Substitute h = 2 into 1 and 2.
2(x – y) = x + y – 1 …3
x + y – 1 = 2x2 – 11y2 …4

From 3, 2x – 2y = x + y – 1
x = 3y – 1…5

Substitute 5 into 4.

3y – 1 + y – 1 = 2(3y – 1)2 – 11y2

4y – 2 = 2(9y2 – 6y + 1) – 11y2

4y – 2 = 18y2 – 12y + 2 – 11y2
7y2 – 16y + 4 = 0

(7y – 2)(y – 2) = 0

y–2=0
2 y=2
y = 7

Substitute y = 2 into 5.
7

x = 3( 2 ) – 1
7

= – 17

Substitute y = 2 into 5.
x = 3(2) – 1
=5

Thus, x = – 1 , y = 2 are the solutions to these simultaneous equations.
7 7

31

5. Let x represents fixed salary as a sales officer, y represents house rental and z
represents online sales.

x + y + z = 20 000 …1
x + 500 = 2(y + z) …2
x + z = 2y …3

From 1, z = −x – y + 20 000…4

From 2, x + 500 = 2y + 2z
x – 2y – 2z = –500 …5

From 3, x – 2y + z = 0…6

Substitute 4 into 5.
x – 2y – 2(−x – y + 20 000) = −500

x – 2y + 2x + 2y – 40 000 = −500
3x = 39 500
x = 13 166.67

Substitute 4 into 6.
x − 2y + (−x – y + 20 000) = 0

3y = 20 000
y = 6 666.67

Substitute x = 13 166.67 and y = 6 666.67 into 4.
z = −x – y + 20 000
= −13 166.67 – 6 666.67 + 20 000
= 166.66

Thus, Raju's income from fixed salary as sales officer, house rental and online sales are
RM13 166.67, RM6 666.67 and RM166.66 respectively.

6. q2 + (2q – 1)2 = p2 …1

p + q + (2q – 1) = 40…2
From 1: q2 + 4q2 – 4q + 1 = p2
5q2 – 4q + 1 = p2…3

p + 3q = 41

Substitute 4 into 3, p = 41 – 3q…4

5q2 – 4q + 1 = (41 – 3q)2

5q2 – 4q + 1 = 1 681 – 246q + 9q2

4q2 – 242q + 1 680 = 0

2(2q – 105)(q – 8) = 0

q= 105 , q = 8
2

32

SGubanstitkuatne q = 8 adnadn q = 105 kinetdoalam, 4,
2
( )p = 41 – 3(8) , 105
p= 41 – 3 2

= 17 = – 2233 (Ignore)

Thus, the sides length of the land are 8 m, 15 m and 17 m.

7. Gradient of straight line = 3 – (–3)
2–0

=3

Equation of straight line: y = 3x − 3 …1

Equation of the curve: x2 + y2 − 27x + 41 = 0 …2

Substitute 1 into 2.

x2 + (3x − 3)2 − 27x + 41 = 0

x2 + 9x2 − 18x + 9 − 27x + 41 = 0

10x2 − 45x + 50 = 0

2x2 − 9x + 10 = 0

(2x − 5)(x − 2) = 0

2x – 5 = 0 or x – 2 = 0

x= 5 x=2
2

Substitute x = 5 into 1.
2
y = 3(25 ) − 3

= 9
2

Substitute x = 2 into 1.
y = 3(2) − 3

= 3

Yes, the line intersect the curve at another point, which is x = 5 , y = 9 .
2 2

8. 3x + y + x + y = 24 …1

(x + y)2 = (3x)2 + y2 …2

From 1, 4x + 2y = 24
2x + y = 12

y = 12 – 2x…3

From 2, x2 + 2xy + y2 = 9x2 + y2
8x2 – 2xy = 0…4

33

Substitute 3 into 4.

8x2 – 2x(12 – 2x) = 0
8x2 – 24x + 4x2 = 0
12x2 – 24x = 0
x2 – 2x = 0

x(x – 2) = 0
x = 0 or x = 2

Substitute x = 0 into 3.
y = 12 – 2(0)
= 12

Substitute x = 2 into 3.
y = 12 – 2(2)
= 12 – 4

=8

Length of wood = 8 cm, Width of wood = 3 × 2 = 6 cm
Thus, the area of the wood is 8 × 6 = 48 cm2.

9. Let x represents the length of the room and y represents the width of the room.
(x − 2)(y − 2) = 8.75 …1

2(x − 2) + 2(y − 2) = 12 …2

From 1, xy – 2x – 2y + 4 = 8.75
xy – 2x – 2y = 4.75…3

From 2, 2x – 4 + 2y – 4 = 12
2x + 2y = 20
x + y = 10
x = 10 − y…4

Substitute 4 into 3. …5

(10 − y)y − 2(10 − y) − 2y = 4.75
10y − y2 −20 + 2y − 2y = 4.75

y2 − 10y + 24.75 = 0

5 × 4: 4y2 – 40y + 99 = 0

(2y – 11)(2y – 9) = 0

2y – 11 = 0 or 2y – 9 = 0
11 9
y = 2 y= 2

SGuabnstiktuatne y = 11 into 4.
2

x = 10 − 11
= 9 2
2

34

Substitute y = 9 into 4.
2

x = 10 − 9
2

= 11
2

Thus, the length and the width of the room is 5.5 m and 4.5 m respectively.

( )10. 22
Length of arc STR, s = 14x 7
= 44x

28xy = 224 …1
28x + 2y + 44x = 72 …2

From 2, 72x + 2y = 72
36x + y = 36
y = 36 – 36x…3

Substitute 3 into 1.

28x(36 − 36x) = 224

1 008x − 1 008x2 = 224

1 008x2 − 1 008x + 224 = 0

9x2 − 9x + 2 = 0

(3x – 2)(3x −1) = 0

3x – 2 = 0 or 3x – 1 = 0

x = 2 x= 1
3 3
2
Substitute x = 3 into 3.

( )y = 36 − 36 2
3

= 12

Substitute x = 1 into 3.
3

( )y = 36 − 36 1
3

= 24

Thus, x = 2 , y = 12 and x = 1 , y = 24.
3 3

35

11. Area of the mural: 7xy = 28 …1

Perimeter ABCDE: 7x + 2y + 22 × 7x = 26
7 2

7x + 2y + 11x = 26

18x + 2y = 26

y= 26 – 18x …2
2

Substitute 2 into 1.

( )7x
26 – 18x = 28
2

91x – 63x2 = 28

63x2 – 91x + 28 = 0

7(9x – 4)(x – 1) = 0
4
x= 9 or x = 1

when x= 4 , the diameter of the semicircle is 28 m and the radius is 14 m.
9 9 9

When x = 1, the diameter of the semicircle is 7 m and the radius is 7 m.
2

36

CHAPTER 4 INDICES, SURDS AND LOGARITHMS

Inquiry 1 (Page 90)

Index law Example

am × an = am + n t2 × t3 = t2 + 3 = t5

am ÷ an = am – n s8 ÷ s3 = s8 – 3 = s5

a0 = 1 u0 = 1

(am)n = amn (p9)2 = p18
n!  a =
am × bm = 1 3!  c 1
n! am =
an = an = c3
bn =
a–n (ab)m k4 × h4 = (kh)4

m 2!  y5 5

an = y2

( a )n ( )r6 = r6
b s
s6

1 d–4 = 1
an d4

1

Inquiry 2 (Page 90)

1. Index law:
• am × an = am + n
• am ÷ an = am – n
• (am)n = amn
• (am × an)p = amp × anp

1

• an

Self Practice 4.1 (Page 92)

1. (a) 53x × 5x = 53x + x – (–x)
5–x = 55x

(b) 7b – 2 – 7b = 7b × 7–2 – 7b
7b + 3 7b × 73
7b(7–2 – 1)
= 7b × 73

= 7–3(7–2 – 1)

= 7–3 + (–2) – 7–3

= 7–5 – 7–3
1 1
= 75 – 73

(c) 9a – 3 + 9a + 4 = 9a × 9–3 + 9a × 94
81 92
9a(9–3 + 94)
= 92

= 9a(9–5 + 92)

(d) c4d3 × c3d5 = c4 + 3d3 + 5
= c7d8

(e) (xy2)3 × x3y5 = x3y6 × x3y5
= x3 + 3y6 + 5
= x6y11

(f) (7x–1)2 × (49–2xy)3 = 49x–2 × 49–6x3y3

= 491 + x(–6) –2 + 3y3

= 49–5xy3

= xy3
495
27x6y3 × x12
(g) (3x2y)3 × (x3)4 ÷ x16y2 = x16y2

= 27x6 + 12 – 16y3 – 2

= 27x2y

(h) (p2q–1)5 × q8 = p10q–5 × q8

= p10q–5 + 8

= p10q3

2

(i) (pq5)4 × p3 = p4q20 × p3
= p4 + 3q20
= p7q20

(j) (49–2xy)3 ÷ (7xy)–2 = 49–6x3y3
7–2x–2y–2
= 7 x y2(–6) – (–2) 3 – (–2) 3 – (–2)

= 7–10x5y5

= x5y5
710
20x–7y2
(k) 20x–7y2 ÷ 4x3y– 4 = 4x3y– 4

= 5x–7 – 3y2 – (– 4)

= 5y6
x10
6a7b–2
(l) 6a7b–2 ÷ 36a3b– 4 = 36a3b– 4

= a b7 – 3 –2 – (– 4)
6
a4b2
= 6

2. 1 × 2a– 12 = 1 + (– 21 )

(a) a3 2a 3 1
2a– 6
=

= 2
1
a6
= 4a3 ÷ a– 53
(b) 4a3 = 4a3 – (– 35 )
3
a– 5

= 4a 158

(c) 5!  a7 × 4!  a–9 = a 75 × a– 9
4

= a  7 + (– 94 )
5

= a– 1270

= 1

17
a20
( ) ( ) ( ) ( )(d)
a– 3 1 + 3a– 32 – 3a– 25 = a– 23 × 1 + a– 23 × 3a– 23 – a– 32 × 3a– 52
2
a2 a2

= a– 23 + 1 + 3a– 32 – 3 – 3a– 23 – 5
2 2 2

= a–1 + 3a–3 – 3a–4

= 1 + 3 – 3
a a3 a4
3. (a) 43a – 2 = 43a × 4–2

= 43a
42
64a
= 16

3

(b) 92a + 2 = 92a × 92
= 81a × 81
= 81(81a)

(c) 73a – 4 = 73a × 7–4
73a
= 74

= 343a
2 401

4. 4x + 2 + 4x + 1 + 4x = (4x × 42) + (4x × 41) + 4x
= 4x(42 + 41 + 1)
= 4x(16 + 4 + 1)
= 4x(21)

Since 21 is a multiple of 7, thus 4x + 2 + 4x + 1 + 4x is divisible by 7 for all positive integer x.

Self Practice 4.2 (Page 94)

1. (a) 4x – 1 = 8x + 3

2 = 22(x – 1) 3(x + 3)

2(x – 1) = 3(x + 3)
2x – 2 = 3x + 9

x = –11

(b) 3x + 3 – 3x + 2 = 2

(3x × 33) – (3x × 32) = 2

3x(33 – 32) = 2

3x(18) = 2
2
3x = 18

3x = 1
9
3x = 3–2

42x x = –2
64
(c) 8x – 3 =

23(x – 3) = 22(2x)
26
24x
23x – 9 = 26

23x – 9 = 24x – 6

3x – 9 = 4x – 6

x = –3

2. (a) h = 10 × (0.9)l
h = 10 × (0.9)0

= 10 cm
(b) h = 10 × (0.9)10

= 3.4868 cm

4

Intensive Practice 4.1 (Page 94)

1. (a) y3(3zx)2 = y332z2x2
9x3 9x3
= y3z2x2 – 3

= y3z2
x
z4yx2
(b) zx y2 = z4 – 1y1 – 2x2 – 1
= z3y–1x

= xz3
y

(c) [(xy)5 × 2xy3]2 = [x5y5 × 2xy3]2
= x10y10 × 4x2y6
= 4x10 + 2y10 + 6
= 4x12y16

(d) (ef 2)3 ÷ (e–2f 2) = e3f 6
e–2f 2

= e3 – f(–2) 6 – 2

= e5f 4

(e) 4.2x4y14 ÷ 0.6x9y5 = 4.2x4y14
0.6x9y5
= 7x4 – 9y14 – 5

= 7x–5y9

= 7y9
x5

(f) (7x–1)2 × (49–2xy)3 ÷ (7xy)–2 = 72x−2 × 7−12x3y3 ÷ 7−2x−2y−2

= 72x –2 × 7–12x3y3
7–2x–2y–2

= 7 x y2 + (−12) − (−2) −2 + 3 − (−2) 3 − (−2)
= 7–8x3y5

= x3y5
78
2. 2x – 2 = 2 (16)
2x – 2 = 32
2x – 2 = 25

x–2=5

x = 7

3. 25x – 53x – 4 = 0
25x = 53x – 4
52x = 53x – 4

2x = 3x – 4

–x = – 4

x=4

5

4. 4(2m + 1) – 16m = 0
4(2m + 1) = 16m
22(2m + 1) = (24)m

2 + m + 1 = 4m

–3m = –3

m=1

5.

START

Solve Solve Solve
2 ÷ 4x – 1 = 162x
3n – 2 × 27n = 1 23x – 5 = 1 1
81 4x +

1 –  2
2 3
1
Solve 2 Solve Solve
4x + 3 – 4x + 2 = 3
25x + 2 = 1 5n + 1 – 5n + 5n – 1
625x = 105

2

Solve Solve Solve
42x – 1 = 64x 324x = 48x + 6 2x + 4 – 2x + 3 = 1

–1 –3

Solve Solve FINISH
16(3n – 1) = 27n 162x – 3 = 84x

6. (a) Let B = 300(3t)

The amount of bacteria after 9 minutes:
B = 300(39)

= 5 904 900

(b) The time taken, in minutes, for the amount of bacteria to be 72 900
B = 300(3t)

72 900 = 300(3t)
3t = 243
3t = 35

t = 5 minutes

6

( )7.P=A 1 + k t, dwehnegraenAA==3030mjiulltiao,nk, =k =0.0.30,3t, =t =20250050– –2021071=7 =3333
100
( )P 3
= 30 000 000 1 + 100 33

= 79 570 057

8. P = f(1 + r)t , wdehnegraenf f==202000000,0r, =r =0.100.1,0t,=t 1=010

( )P = 1 + 10 10
20 000 100

= RM51 874.85

Inquiry 3 (Page 96)

3. Let a = 2, b = 3 and c is the hypotenuse.

c = ! 22 + 32

= ! 13 2
! 13
k cooss q =

Inquiry 4 (Page 96) Irrational numbers
Rational numbers
Surd Not surd

0.333333… ! 2 π

0.141414… 3! 121 e = 2.71828182845…

3.4566666… ! 3
5.8686 4
0.5
0.175 ! 7
9

!5  3343

!3 1
8

1.234567…

0.5555…

7

Mind Challenge (Page 98)

N= 224
495

0.45252

495 ) 2 2 4 0

– 1 9 8 0
2 6 0 0

– 2 4 7 5
1 2 5 0

– 9 9 0
2 6 0 0

– 2 4 7 5
1 2 5 0

– 9 9 0
2 6 0 0

Self Practice 4.3 (Page 99)

1. (a) Let N = 0.787878…1
1 × 100: 100N = 78.7878…2

2 – 1: 99N = 78
78
N= 99

MTahkuas, 0.787878… = 26
33

(b) Let A = 3.57575757

A=3+N
N = 0.57575757…1
Assume

1 × 100: 100N = 57.575757…2

2 – 1: 99N = 57

N= 57
99
19
= 33

A=3+ 19
33
19
= 3 33

MTahkuas, 3.57575757… = 3 19
33

(c) Let N = 0.345345345…1
1 × 1 000: 1 000N = 345.345345…2

2 – 1: 999N = 345

N= 345
999
115
= 333

MTahkuas, 0.345345345… = 115
333
8

(d) Let A = 13.567567567

A = 13 + N
Assume N = 0.567567567…1
1 000N = 567.567567…2
2 – 1: 999N = 567

N = 567
999
21
= 37

A = 13 + 21
37
21
= 13 37

MThauksa, 13.567567567… = 13 21
(a) 3! 127 = 5.026525695… 37
2.

3! 127 is a surd because it is a non-recurring decimal.

(b) 4! 1125 = 5.791460926…
4! 1125 iasdalsauhrdsubredckauersaeniat ims eanngohna-srielckuarnripnegrpduelcuimhaanl.tidak berulang.

(c) 6! 64 = 0.6666666…
729

6! 64 iasdnaolatha bsukrdanbescuarudskeeirtainsaamreecnugrhriansgilkdaencimpearlp.uluhan berulang.
729

(d) 7! 79 = 0.706743939…
897

7! 79 aisdalsauhrdsubredckauersaeniat ims eanngohna-srielkcuanrripnegrpduelcuihmaanl.tidak berulang.
897

Inquiry 5 (Page 99)

a b (a × b) !a !b Value of ! (a × b) Value of
! a × !  b ! (a × b)

25 10 ! 2 ! 5 3.162 ! 10 3.162

34 12 ! 3 ! 4 3.46410 ! 12 3.46410

57 35 ! 5 ! 7 5.91607 ! 35 5.91607

11 13 35 ! 11 ! 12 11.9583 ! 143 11.9583

a b (a ÷ b) !a !b Value of ! (a ÷ b) Value of
! a ÷ !  b ! (a ÷ b)

10 5 2 ! 10 ! 5 1.414 !2 1.414

15 3 5 ! 15 ! 3 1.732 !3 1.732

35 5 7 ! 35 ! 5 2.6457 !7 2.6457

66 11 6 ! 66 ! 11 2.4495 ! 11 2.4495

9

Self Practice 4.4 (Page 101) (b) !  3 × !  5 = !  15
1. (a) !  2 × !  3 = !  6 (d) !  5 × !  6 = !  30

(c) !  3 × !  3 = !  9

(e) !  8 = ! 8 (f) !!  138 = ! 138 = !  6
!3 3

(g) !  20 = ! 250 = !  4 (h) !  5 × !  6 = ! 330 = !  10
!5 !3

Inquiry 6 (Page 101)

3. Surds that cannot be simplified
Surds that can be simplified

! 1 , !  4 , !  8 , !  9 , !  12 , !  16 , !  18 , !  20 , ! 2 ,!  3 ,!  5 ,!  6 ,!  7 ,!  10 ,!  11 ,!  13 ,!  14 ,

! 24 , !  25 , !  27 , !  28 , !  32 , !  36 , !  40 , ! 15 , !  17 , !  19 , !  21 , !  22 , !  23 , !  26 ,

! 44 , !  45 , !  48 , !  49 , !  50 ! 29 , ! 30 , !  31 , !  33 , !  34 , !  35 , !  37 ,

! 38 , !  39 , !  41 , !  42 , !  43 , !  46 , !  47

4. ! 90 = ! 9 × 10 3! 2 × 2! 2 = 6 × ! 2 × ! 2
= 3! 10 = 12

Self Practice 4.5 (Page 101) (! 16 ! 36 )2 = (! 576 )2 3
1. ! 5 ! 7 = ! 5 × 7 3
3 = 576 3
= ! 35 3

! 260 = ! 4 × 65
= 2! 65

4! 7 × 5! 7 = 20! 7 ×! 7 4! 8 = 2! 8
2! 4 4
= 2! 2
= 140

!! 18=  138 ! 75 = ! 735
! 3
!3 = ! 25
= ! 6

= 5

! 360!!  327 = 5  237 (! 81 )2 = 81
= 5! 9
= 15 3

10

2. (a) ! 12 = ! 4 × 3 (b) ! 27 = ! 9 × 3
= 2!  3 = 3!  3

(c) ! 28 = ! 4 × 7 (d) ! 32 = ! 16 × 2
= 2!  7 = 4!  2

(e) ! 45 = ! 9 × 5 (f) ! 48 = ! 16 × 3
= 3!  5 = 4!  3

(g) ! 54 = ! 9 × 6 (h) ! 108 = ! 36 × 3
= 3! 6 = 6!  3

Inquiry 7 (Page 102)

5. Calculation steps:
(a) Collect all the same surds
(b) Perform addition and subtraction as other algebraic solutions.

SLealftiPhrDaicrtic4e.64.6 ((HPaglaem10a3n) 103)

1. (a) 3!  5 + 5!  5 (b) 7!  5 + 5!  5
= !  5 (3 + 5) = !  5 (7 + 5)
= 8!  5 = 12!  5

(c) 7!  7 – 5!  7 (d) !  6 (3!  6 – 5!  6 )
= !  7 (7 – 5) = 3(!  6 ×!  6 ) – 5(!  6 ×!  6 )
= 2!  7
= 3(6) – 5(6)

(e) !  5 (4 + 5!  5 ) = –12
= 4!  5 + 5(!  5 ×!  5 )
(f) !  7 (3 – 5!  7 )
= 4!  5 + 25 = 3!  7 – 5(!  7 ×!  7 )

= 3!  7 – 35

(g) (4 + 5!  3 )(3 + 5!  3 ) = (4 × 3) + (4 × 5!  3 ) + (3 × 5!  3 ) + (5!  3 × 5!  3 )

= 12 + 20!  3 + 15!  3 + 75
= 87 + 35!  3

(h) (7 – 5!  7 )(3 + 5!  7 ) = (7 × 3) + (7 × 5!  7 ) – (3 × 5!  7 ) – (5!  7 × 5!  7 )

= 21 + 35!  7 – 15!  7 – 175
= 20!  7 – 154

(i) (9 + 5!  4 )(3 – 5!  4 ) = (9 × 3) – (9 × 5!  4 ) + (3 × 5!  4 ) – (5!  4 × 5!  4 )

= 27 – 45!  4 + 15!  4 – 100

= –133

2. (a) 5!  80 = 5!  16 × 5 , 2!  58 = 2!  29 × 2 , 9!  45 = 9!  9 × 5
= 27!  5
= 20!  5 = 2!  58
SNuortdstiamkilsaerrsuuprads

11

(b) 3!  3 , 4!  12 = 4!  4 × 3 , 5!  27 = 5!  9 × 3
= 8!  3 = 15!  3
SSimuridlarsesruurpdas
7!  5 , –7!  5
(c) 2!  125 = 2!  25 × 5 ,

= 10!  5
SSimuridlasresruurpdas

(d) 2!  12 = 2!  4 × 3 , 9!  24 = 9!  4 × 6 , 8!  5
= 18!  6
= 4!  3
NSoutrsdimtaiklasresruurpdas

(e) 3!  27 = 3!  9 × 3 , –3!  27 = –3!  9 × 3 , –!  3
= –9!  3
= 9!  3
SSimuridlasresruurpdas

Mind Challenge (Page 105)
1+!3

Self Practice 4.7 (Page 106)

1. (a) 2 = 2 × ! 5 (b) 7 = 7 × ! 2
! 5 ! 5 ! 5 !2 ! 2 ! 2

= 2! 5 = 7! 2
5 2

(c) ! 2 = ! 2 × ! 5 (d) 2!!  312 = 2!!  312 × 22!!  1122
!5 ! 5 ! 5

= ! 10 = 2! 36
5 48
12
= 48

= 1
4

(e) 1 + ! 3 = 1 + ! 3 × ! 12 (f) 35 +– !!  52 = 35 + ! 2 × 5 + ! 5
! 12 ! 12 ! 12 – ! 5 5 + ! 5

= ! 12 + ! 36 = 15 + 3! 5 + 5! 2 + ! 5 ! 2
12 25 – 5

= 2! 3 + 6 = 15 + 3! 5 + 5! 2 + ! 10
12 20

= ! 3 + 1
6 2

12

(g) 6 – ! 3 = 6 – ! 3 × 9 + ! 12 = 54 + 6! 12 – 9! 3 + ! 3 ! 12
9 – ! 12 9 – ! 12 9 + ! 12 81 – 12

= 54 + 6! 12 – 9! 3 – ! 36
81 – 12

= 54 + 12! 3 – 9! 3 –6
69

= 48 + 3! 3
69

= 16 + ! 3
23

3 + ! 2 4 – ! 3 3 + ! 2 5 + ! 2 4 – ! 3 7 – ! 3
5 – ! 2 7 + ! 3 5 – ! 2 5 + ! 2 7 + ! 3 7 – ! 3
( ) ( ) ( ) ( )(h)+ = × + ×

= 15 + 3! 2 + 5! 2 + 2 + 28 – 4! 3 – 7! 3 + 3
25 + 5! 2 – 5! 2 – 2 49 – 7! 3 – 7! 3 – 3

= 17 + 8! 2 + 31 – 11! 3
23 46

= 2(17 + 8! 2 ) + 31 – 11! 3

46

= 34 + 16! 2 + 31 – 11! 3
46

= 65 + 16! 2 – 11! 3
46

7 – ! 5 6 + ! 3 7 – ! 5 5 – ! 5 6 + ! 3 6 + ! 3
5 + ! 5 6 – ! 3 5 + ! 5 5 – ! 5 6 – ! 3 6 + ! 3
( ) ( ) ( ) ( )(i)– = × – ×

= 35 – 7! 5 – 5! 5 + 5 + 36 + 6! 3 + 6! 3 + 3
25 + 5! 5 – 5! 5 – 5 36 + 6! 3 – 6! 3 – 3

= 40 – 12! 5 + 39 + 12! 3
20 33

= 4(10 – 3! 5 ) + 3(13 + 4! 3 )

20 33

= 10 – 3! 5 + 13 + 4! 3
5 11

= 11(10 – 3! 5 ) – 5(13 + 4! 3 )

55

= 110 – 33! 5 – 65 – 20! 3
55

= 45 – 33! 5 – 20! 3
55

13

Self Practice 4.8 (Page 106)

1. h = sin 60° × 3! 3 A

= ! 3 × 3! 3 3ͱස3සcm
2
9
= 2 B 60° h
x C
x = cos 60° × 3! 3
1 4ͱස3සcm
= 2 × 3! 3

= 3! 3
2

AC2 = h2 + (4! 3 – x)2

( ) ( )=92+ 4! 3 – 3! 3 2
2 2
81 75
= 4 + 4

= 39

AC = ! 39 cm

2. (a) Area = 1 (5 – 2! 2 )(5 + 2! 2 )
2
1
= 2 (25 + 10! 2 – 10! 2 – 8)

= 1 × 17
2
17
= 2 cm2

(b) AC2 = (5 – 2! 2 )2 + (5 + 2! 2 )2
= (25 – 20! 2 + 8) + (25 + 20! 2 + 8)

= 50 + 16
= 66

A C = ! 66 cm
3. 2 + 3! y = 6! 3 + 5

3! y = 6! 3 + 3

! y = 6! 3 + 3
3

( )y =
6! 3 + 3 2
3

= 108 + 36! 3 +9
9

= 117 + 36! 3
9

= 13 + 4! 3

14

4. (a) ! 2 –7x + 2x = 0

! 2 –7x = –2x

KSquuaasraedbuoatkhasnidkeesdua-dua belah
2 – 7x = 4x2

4x2 + 7x – 2 = 0

(4x – 1)(x + 2) = 0
1
x= 4 ((AIgbnaoikrea)n)  oatrau  x = −2

MThakuas, x = –2.

(b) ! 2x + 1 + ! 2x – 1 = 2

! 2x + 1 = 2 – ! 2x – 1

KSuqausaaredbuoakthansidkeesdua-dua belah
2x + 1 = 4 − 4! 2x – 1 + 2x − 1

4! 2x – 1 = 2

! 2x – 1 = 1
2

KSquuaasraedbuoatkhasnidkeesdua-dua belah
1
2x – 1 = 4

2x = 5
4
5
x= 8

(c) ! 4x + 3 – ! 4x – 1 = 2
! 4x + 3 = 2 + ! 4x – 1

KSquuaasraedbuoatkhasnidkeesdua-dua belah
4x + 3 = 4 + 4! 4x – 1 + 4x − 1
4! 4x – 1 = 0

KSquuaasraedbuoatkhasnidkeesdua-dua belah

4x – 1 = 0

4x = 1

x= 1
4

Intensive Practice 4.2 (Page 108)

1. (a) ! 5 × ! 11 = ! 55 (b) ! 7 × ! 10 = ! 70

!(c)! 27=  2178 !(d)! 48 =  488
! 18 !8

= ! 3 = ! 6
2

15

2. (a) ! 24 = ! 4 × ! 6 (b) ! 162 = ! 81 × ! 2
= 2! 6 = 9! 2

! 54!(c)=  534 ( )(d) 2!3 6 2 = 4 ×9 6
!3
= ! 18 24
= 9

= ! 9 × ! 2 = 8
3
= 3! 2 4 ! 4
= 3

3. (a) 3! 10 + 5! 10 = 8! 10 (b) 6! 11 – ! 11 = 5! 11

(c) 13! 13 – 2! 13 = 11! 13 (d) 2! 45 + ! 20 = 2! 9 × 5 + ! 4 × 5
= 6! 5 + 2! 5
= 8! 5

(e) 3! 27 – ! 72 = 3! 9 × 3 – ! 72 (f) ! 18 + ! 27 = ! 9 × 2 + ! 9 × 3
= 3! 9 × 3 – ! 9 × 8 = 3! 2 + 3! 3
= 9! 3 – 3! 4 × 2
= 9! 3 – 6! 2 (h) ! 72 × 4! 15 = 4! 1 080
= 4! 36 × 30
(g) 3! 15 × 7! 5 = 21! 15 × 5 = 24! 30
= 21! 75
= 21! 25 × 3 (j) ! 7 (3 + 7! 7 ) = 3! 7 + 49
= 105! 3
(l) (3 + 3! 7 )(3 + 5! 7 ) = 9 + 15! 7 + 9! 7 + 105
(i) ! 4 (2! 3 ) – 5! 3 = 2! 12 – 5! 3

= 2! 4 × 3 – 5! 3
= 4! 3 – 5! 3
= –! 3

(k) ! 5 (7 – 5! 5 ) = 7! 5 – 25

= 114 + 24! 7

(m) (7 + 5! 7 )(3 – 5! 7 ) (n) (7 – 5! 5 )(3 – 5! 5 )

= 21 – 35! 7 + 15! 7 – 175 = 21 – 35! 5 – 15! 5 + 125

= –154 – 20! 7 = 146 – 50! 5

! 112!(o)=  1172 !(p) !! 1 1028 =  11028
!7

= ! 16 = ! 1
9
=4 1
= 3

16

!(q) ! 88 = 1  8118 !(r) 9! 20 = 3  250
2! 11 2 3! 5
1
= 2 ! 8 = 3! 4

= 1 ! 4 × 2 =6
2

= ! 2

4. Given A = 3! 5 + 7! 3 , B = 2! 5 – 7! 7 , C = 2! 3 – 9! 8

(a) A + B = 3! 5 + 7! 3 + 2! 5 – 7! 7
= 5! 5 + 7! 3 – 7! 7

(b) A – C = (3! 5 + 7! 3 ) – (2! 3 – 9! 8 )

= 3! 5 + 5! 3 + 9! 8
= 3! 5 + 5! 3 + 18! 2

(c) 3A + 2B = 3(3! 5 + 7! 3 ) + 2(2! 5 – 7! 7 )

= 9! 5 + 21! 3 + 4! 5 – 14! 7
= 13! 5 + 21! 3 – 14! 7

(d) 3A + B – 2C = 3(3! 5 + 7! 3 ) + (2! 5 – 7! 7 ) – 2(2! 3 – 9! 8 )

= 9! 5 + 21! 3 + 2! 5 – 7! 7 – 4! 3 + 18! 8
= 11! 5 + 17! 3 – 7! 7 + 18! 8
= 11! 5 + 17! 3 – 7! 7 + 36! 2

5. (a) 2 = 2 × ! 5 (b) 4 = 4 × 3 + ! 5
! 5 ! 5 ! 5 3 – ! 5 3 – ! 5 3 + ! 5

= 2! 5 = 12 + 4! 5
5 9–5

= 3 + ! 5

(c) 4 = 3 4 × 3 + 3! 5 (d) 5 = 5 × 2! 3 + ! 2
3 – 3! 5 – 3! 5 3 + 3! 5 2! 3 – ! 2 2! 3 – ! 2 2! 3 + ! 2

= 12 + 12! 5 = 10! 3 + 5! 2
9 – 45 12 – 2

= 12 + 12! 5 = 10! 3 + 5! 2
–36 10
– (1 )
= + ! 5 = ! 3 + ! 2
3 2

17

(e) 4 + ! 5 = 4 + ! 5 × 3 + ! 5 (f) ! 3 – ! 7 = ! 3 – ! 7 × ! 3 – ! 7
3 – ! 5 3 – ! 5 3 + ! 5 ! 3 + ! 7 ! 3 + ! 7 ! 3 – ! 7

= 12 + 4! 5 + 3! 5 +5 = 3 – ! 3 ! 7 – ! 3 ! 7 + 7
9–5 3–7

= 17 + 7! 5 = 10 – 2! 21
4 – 4

= –5 + ! 21
2
6. (a) 1 + 1 = 1 – ! 3 + 1 + ! 3
1 + ! 3 1 – ! 3
(1 + ! 3 )(1 – ! 3 )
2
= 1–3

= −1

(b) !7 2 + ! 7 1 = 2(! 7 – ! 2 ) + ! 7 + ! 2
+ ! 2 – ! 2 (! 7 + ! 2 )(! 7 – ! 2 )

= 2! 7 – 2! 2 + ! 7 + ! 2
7–2

= 3! 7 – ! 2
5

(c) 2 + 4 1 = 2(4 + ! 3  ) + 4 – ! 3
4 – ! 3 + ! 3
16 – 3

= 8 + 2! 3 + 4 – ! 3
13

= 12 + ! 3
13
7. (! 5 + ! 2 )p = 8 + ! 10

p = 8 + ! 10 × ! 5 – ! 2
! 5 + ! 2 ! 5 – ! 2

= 8! 5 – 8! 2 + ! 50 – ! 20
5–2

= 8! 5 – 8! 2 + 5! 2 – 2! 5
5 –2

6! 5 – 3! 2
3
( )=
= 2! 5 – ! 2 cm

18

8. (a) tan x = 3 + ! 2
1 + 2! 2

= 3 + ! 2 × 1 – 2! 2
1 + 2! 2 1 – 2! 2

= 3 – 6! 2 + ! 2 –4
1–8

= –1 – 5! 2
–7

= 1 + 5! 2
7
1
(b) Area = 2 × (3 + !2) × (1 + 2! 2 )

= 1 × (3 + 6! 2 + ! 2 + 4)
2
1 (7 7! 2 )
= 2 × +

( )=
7 + 7! 2 cm2
2

Inquiry 8 (Page 109)

7. The logarithmic value of a negative number and zero cannot be determined.
8. (a) loga 1 = 0 is valid.

(b) loga a = 1 is valid.

Inquiry 9 (Page 111)

x –3 –2 –1 0 1 2 3

y 1 1 1 1 2 4 8
8 4 2

x –  1 –  1 –  1 1 2 4 8
8 4 2

y 3210123

y

8 x
6
4
2

–6 –4 –2–20 2 4 6 8 10
–4

19

Self Practice 4.9 (Page 113)

1. (a) 34 = 81 (b) 27 = 128
log3 81 = 4 log2 128 = 7

(c) 53 = 125 (d) 63 = 216
log6 216 = 3
log5 125 = 3
(b) log10 0.0001 = −4
2. (a) log10 10 000 = 4 10−4 = 0.0001
104 = 10 000
(d) log4 64 = 3
(c) log2 128 = 7 43 = 64
27 = 128

3. (a) log10 9 = 0.9542 (b) log10 99 = 1.9956

( )(c) log10 5 3 = –0.2375 (d) log2 64 = x
6 2x = 64
2x = 26
(e) log3 81 = x x=6
3x = 81
3x = 34 (f) log4 256 = x
4x = 256
x = 4 4x = 44
x=4

(g) log10 100 000 = x
10x = 100 000
10x = 105

x=5

4. (a) log2 x = 5 (b) log8 x = 3 (c) log2 x = 8
25 = x 83 = x 28 = x

x = 32 x = 512 x = 256

5. (a) antilog 2.1423 = 138.7714 (b) antilog 1.3923 = 24.6774
(c) antilog 3.7457 = 5568.0099 (d) antilog (–3.3923) = 0.0004052
(e) antilog (–2.5676) = 0.0027064 (f) antilog (– 4.5555) = 0.000027829

Inquiry 10 (Page 113)

4. JIifka,ax, axndaynayreiaploashitpivoesiatinfddan≠a1,≠th1e,nmaka

(a) loga xy = loga x + loga y  ((HPurokduumcthlaswil) darab)
x
(b) loga y = loga x – loga y  ((HDuivkiusmionhalaswil )darab)

(c) loga xn = n loga x fuonrtuaknyserbealrannugmnboemr nbo(rPonwyaetralnaw  ()Hukum kuasa)

Self Practice 4.10 (Page 115)

1. (a) log7 1 1 = log7 5 (b) log7 28 = log7 (7 × 4)
4 4
= log7 5 − log7 4 = log7 7 + log7 4
= 0.827 – 0.712 = 1 + 0.712
= 1.712
= 0.115 20

(c) log7 100 = log7 (25 × 4) ( )(d) log7 0.25 = log7 1
4
= log7 52 + log7 4 = log7 1 − log7 4
= 2 log7 5 + log7 4 = −0.712
= 2(0.827) + 0.712
= 2.366

2. (a) log3 21 + log3 18 – log3 14 = log3 (21 × 18 ÷ 14)
= log3 27
= log3 33
= 3 log3 3
=3
1 1
(b) 2 log4 2 – 2 log4
9 + log4 12 = log4 (22 × 12 ÷ 92)
= log4 16
= log4 42
= 2 log4 4
=2

(c) log2 7 + log2 12 – log2 21 = log2 (7 × 12 ÷ 21)
= log2 4
= l2oglo2g2222
=

=2

Mind Challenge (Page 116)

(a) log10 (–6) is undefined.
(b) log–10 (6) is undefined.

Self Practice 4.11 (Page 116)

1. (a) log2 x + log2 y2 = log2 xy2 logb x – 3 logb y = logb x – logb y3
( )(b) x
(c) log2 x + 3 log2 y = log2 x + log2 y3 y3
= log2 xy3 (d)
1 = logb
2 log4 x + 2 – 3 log4 y

1
= log4 x2 + log4 42 – log4 y3

( )= log4 16! x
y3
(e) log3 m4 + 2 log3 n – log3 m = log3 m4 + log3 n2 – log3 m
m4n2
( )= log3 m

= log3 m3n2

2. (a) log2 10 = log2 (2 × 5) (b) log2 45 = log2 (9 × 5)
= log2 2 + log2 5 = log2 32 + log2 5
= 1 + q = 2 log2 3 + log2 5
= 2p + q

21

(c) log2 ! 15 = log2 1

15 2

= 1 log2 (3 × 5)
2
1 1
= 2 log2 3 + 2 log2 5

= 1 p + 1 q
2 2
1
= 2 (p + q)

Mind Challenge (Page 117)

log5 20 = log10 20 log5 20 = loge 20
log10 5 loge 5

= 1.30103 = 2.99573
0.69897 1.60944
= 1.8614
= 1.8614

Self Practice 4.12 (Page 118)

1. (a) log3 22 = log10 22 (b) log6 1.32 = log10 1.32
log10 3 log10 6

= 1.3424 = 0.1206
0.4771 0.7782
= 2.8137 = 0.1550

(c) log5 18 = log10 18 (d) log4 0.815 = log10 0.815
log10 5 log10 4

= 1.2553 = –0.0888
0.6990 0.6021
= 1.7959 = −0.1475

2. (a) log7 225 = loge 225 (b) log9 324 = loge 324
loge 7 loge 9

= 5.4161 = 5.7807
1.9459 2.1972
= 2.7833 = 2.6309

(c) log20 379 = loge 379
loge 20

= 5.9375
2.9957
= 1.9820

3. (a) log2 9 = log3 9 (b) log9 8 = log3 8
log3 2 log3 9

= log3 32 = log3 23
log3 2 log3 32

= 2 = 3 log3 2
t 2 log3 3
3t
= 2

22

(c) log2 18 = log3 18 (d) log2 9 = log3 9
log3 2 4 4
log3 2

= log3 (9 × 2) = log3 9 – log3 4
log3 2 log3 2

= log3 9 + log3 2 = log3 32 – log3 22
log3 2 log3 2

= 2 log3 3 + log3 2 = 2 log3 3 + 2 log3 2
log3 2 log3 2
2+t 2 – 2t
= t = t

( )(b) m
4. (a) log4 m2n3 = log2 m2n3 log8 m = log2 n2
log2 4 n2 log2 8

= log2 m2 + log2 n3 = log2 m – log2 n2
log2 22 log2 23

= 2 log2 m + 3 log2 n = log2 m – 2 log2 n
2 log2 2 3 log2 2

= 2a + 3b = a – 2b
2 3

(c) logmn 8n = log2 8n
log2 mn

= log2 23 + log2 n
log2 m + log2 n

= 3+ b
a+ b

Self Practice 4.13 (Page 120)

1. (a) 42x – 1 = 7x

(2x – 1) log 4 = x log 7

2x log 4 – log 4 = x log 7

2x log 4 – x log 7 = log 4

x(2 log 4 – log 7) = log 4

x = 2 log 4 7
log 4 – log
0.6021
= 1.2041 – 0.8451

= 1.677

23

(b) 52x − 1 = 79x − 1

(2x – 1) log 5 = (x – 1) log 79

2x log 5 – log 5 = x log 79 – log 79

2x log 5 – x log 79 = log 5 – log 79

x(2 log 5 – log 79) = log 5 – log 79

x= log 5 – log 79
2 log 5 – log 79

= 0.6990 – 1.8976
1.3979 – 1.8976
= 2.399

(c) 73x – 1 = 50x

(3x – 1) log 7 = x log 50

3x log 7 – log 7 = x log 50

3x log 7 – x log 50 = log 7

x(3 log 7 – log 50) = log 7
log 7
x= 3 log 7 – log 50

= 0.8451
2.5353 – 1.6990
= 1.011

2. (a) ln (5x + 2) = 15

loge (5x + 2) = 15
e15 = 5x + 2

3 269017.372 = 5x + 2

5x = 3269017.372 – 2

x= 3269015.372
5
= 653803.075

(b) 30e2x + 2 = 145

e2x + 3 = 145
30
e2x + 3 = 4.8333

ln e2x + 3 = ln 4.8333

(2x + 3) ln e = ln 4.8333

2x = ln 4.8333 – 3

x = –1.4245
2
x = −0.712

(c) 5e3x – 4 = 35

e3x – 4 = 35
5
=7

ln e3x – 4 = ln 7

(3x – 4) ln e = ln 7

3x = ln 7 + 4

x= 5.9460
3
x = 1.982
24

(d) ln (3x – 2) = 4

loge (3x – 2) = 4
e4 = 3x − 2

54.5982 = 3x – 2

3x = 54.5982 + 2

= 56.5982
3
= 18.866

(e) 41 – e2x = 5

e2x = 41 − 5

ln e2x = ln 36

2x ln e = ln 36

x= ln 36
2
3.5835
= 2

x = 1.792

(f) ln (x + 1)2 = 4

(x + 1)2 = e4

(x + 1) = ±e2

x = e2 – 1 x = –e2 – 1

= 7.3891 – 1 = –7.3891 – 1

= 6.3891 = –8.389

( )3. 9n
H = 260 000 8

( )260 0009 n . 300 000
8
( )9 30
8 n . 26

n log 9 . log 30
8 26
30
n . log 26

log 9
8
n . 1.2150

The minimum number of years for the price of the house to exceed RM300 000 is 2 years.

4. S = 2000 (1 + 0.07)n
2000 (1 + 0.07)n . 4 000
(1 + 0.07)n . 2
n log (1 + 0.07) . log 2

n . log 2
log (1 + 0.07)

. 00..0302 1904
. 10.23 81
TBhileanmgiannimtauhmunnmuminbiemruomf yuenatruskfojur mthleahsasviimngpsantoanenxycaeemdeRleMbi4hi0R00Mi4s 10100yeiarlas.h 11 tahun.

25

5. W = 4000(1.1)n
4 000(1.1)n . 5100

(1.1)n . 5100
4000
5100
n log 1.1 . log 4000

n . log 1.275
log 1.1

. 00 ..10045145
n . 2.5483
The number of years for Mr. Chong's money to exceed RM5 100 for the first time is 3 years.

6. P = 760e−0.125h

760e−0.125h = 380

e−0.125h = 380
760
1
−0.125h log e = log 2

h= log 1
2
–0.125 log e
–0.3010
= –0.0543

= 5.543 km

Intensive Practice 4.3 (Page 121)

1. log5 1 = 0 log5 75
log5 7
log7 75 =

= log5 (25 × 3)
log5 7

= log5 25 + log5 3
log5 7

= 2 + 0.683
1.209
= 2.219

( )2.loga45 = loga 45 – loga a3
a3 = loga (32 × 5) – 3 loga a

= 2 loga 3 + loga 5 – 3
= 2x + y − 3

26

3. log4 8 + logr ! r = log2 8 1
log2 4
+ logr r2

= log2 8 + 1 logr r
log2 4 2

= log2 23 + 1 logr r
log2 22 2
3 1
= 2 + 2

=2

4. log12 49 × log64 12 = log 49 × log 12
log16 7 log 12 log 64

log 7
log 16

= log 49 × log 16
log 64 × log 7

= log 72 × log 42
log 43 × log 7
4
= 3

5. log10 x = 2  ,  log10 y = −1
x = 102 y = 10−1

xy – 100y2 = 102 × 10−1 – 100 × (10−1)2

= 102 – 1 – 102 × 10−2

= 10 − 100

= 10 – 1

= 9 (proved)
49
( )6. 10
log5 4.9 = log5

= log5 49 – log5 10
= log5 72 – log5 (2 × 5)
= 2 log5 7 – log5 2 – log55
= 2p – m – 1

7. log2 (2x + 1) – 5 log4 x2 + 4 log2 x = log2 (2x + 1) + log2 x4 – 5 log4 x2

= log2 (2x + 1)x4 – 5 log2 x2
log2 4
5
= log2 (2x5 + x4) − 2 log2 x2

= log2 (2x5 + x4)
5
(x2) 2

2x5 + x4
x5
( )= log2

( )= log2 2+ 1
x

27

8. log2 xy = 2 + 3 log2 x – log2 y
= log2 4 + log2 x3 – log2 y

( )= log2 4x3
y
4x3
xy = y

y2 = 4x2

y = 2x

( ) ( )9.log48b= log2 8b
c c
log2 4
1
= 2 (log2 8 + log2 b – log2 c)

= 1 (3 + x – y)
2
( ) 10. 10–7
(a) 10 log10 P0 = 50

( )log1010–7 =5
P0

10–7 = 105
P0 = 10–7
P0 105

P0 = 10−12 Wwatt

(b) Dishwasher : Hot water pump
62 : 50

31 : 25
( )(c) d = 10 log10 P
P0

( )= 10 log10 100
10–12
= 10 log10 1014
= 140
The power of more than 140 decibels will be painful to the human ears.

11. (a) P = 2 500 000e0.04(0)
= 2 500 000
(b) P = 2 500 000e0.04(10)
= 3 729 561
Thus, the population of that country in 2030 is 3 729 407.
(c) 2 500 000e0.04t . 50 000 000
50 000 000
e0.04t . 2 500 000

e0.04t . 20
0.04t log e . log 20
log 20
t . 0.04 × log e

. 74.8933

= 75

2020 + 75 = 2095. Thus, the population of that country exceeds 50 000 000 in 2095.
28

Self Practice 4.14 (Page 123)

1. A = 1 000 × 20.7n

5 000 = 1 000 × 20.7n

20.7n = 5

log 20.7n = log 5

0.7n log 2 = log 5

n= log 5
0.7 log 2

= 3.3170

Thus, the time taken for the insects to infest an area of 5 000 hectares is 3 weeks and 2 days.

2. (a) I = 32 × 4−0

= 32 amp

(b) (i) I = 32 × 4−1

= 8 amp

(ii) I = 32 × 4−2

= 2 amp

(c) 0.5 = 32 × 4−t

4−t = 0.5
32
4t = 64

log 4t = log 64

t = log 64
log 4
t = 3 seconds

Intensive Practice 4.4 (Page 123)
1. (a) W = 1 000(1.09)5

= 1538.62

Thus, the amount of Mr. Ramasamy's money after 5 years is RM1 538.62.
(b) 1 200 = 1 000(1.09)t

1.09t = 1 200
1 000
1.09t = 1.2

t log 1.09 = log 1.2

t= log 1.2
log 1.09

= 2.1156

Thus, the time needed by Mr. Ramasamy to obtain RM1 200 is 2.1156 years.

2. (a) W(t) = 50 × 2–0.0002(0)

= 50
Thus, the initial mass of the uranium is 50 gram.

29

(b) 8 = 50 × 2–0.0002t
8
2 =–0.0002t 50

2–0.0002t = 0.16

–0.0002t log 2 = log 0.16

t= log 0.16
–0.0002 log 2

= 13219.2810
MThauksa, tmheastaimyeanngeeddipeedrlfuokratnheumntuasksjoisfimuraunraiunmiumtobbeerbaatk8i g8ragmrasmisia1l3ah21193.21891.028y1e0artsa.hun.

3. (a) J = 25 × e0.1t

25 × e0.1t = 50

e0.1t = 50
25
e0.1t = 2

0.1t ln e = ln 2

t= ln 2
0.1
= 10 ln 2
MTahkuas, t = 10 ln 2 hjaomur.s

(b) t = 10 ln 2

= 10 × 0.6931

= 6.931
MTahkuas, t = 6.93 hjaomur.s

LaMtihaastnerPyenPgruacktuihcean (H(Paalagme 1a2n4)124)

1. 42x – 1 + 42x = 4
( )42x 1
4 +1 =4

42x(1.25) = 4

42x = 4
1.25
42x = 3.2

2x log 4 = log 3.2

x= log 3.2
2 log 4

= 0.5051
1.2041
= 0.4194

2. 5n + 1 – 5n + 5n – 1 = 105
5n
5n × 5 – 5n + 5 = 105

( )5n5–1 + 1 = 105
5
5n(4.2) = 105

5n = 105
4.2
= 25

5n = 52

n=2 30