MATEMATIK TG3 - BAB 1

BAB 1 Indeks

Indices

  1.1  Tatatanda Indeks Buku Teks: m.s. 2 – 5

A. Lengkapkan jadual dengan asas dan indeks bagi nombor yang diberi. NOTA
Complete the table with the bases and indices of the given numbers.  SP1.1.1  TP1
Bentuk indeks:
Nombor 1 75 12 Index form:
Number 34 0.26 (–  4)3 5 p8 Indeks/Index
3 p a↓n
Asas 4 7 0.2 – 4 1 8 ↑
Base 5 Asas/Base

Indeks 5632
Index

B. Nyatakan yang berikut dalam bentuk indeks atau pendaraban berulang.
State the following in the index form or repeated multiplication.  SP1.1.1  TP2

1. 46 = 4 × 4 × 4 × 4 × 4 × 4 2. (–7) × (–7) × (–7) = (–7)3

Berulang empat kali
Repeated four times

(a) 54 = 5 × 5 × 5 × 5 3. h5 = h × h × h × h × h 1  4. 1 × 1 × 1 = 13
(b) (–2)3 = (–2) × (–2) × (–2) n n n n
(c) m × m × m × m = m4

C. Tulis setiap nombor dalam bentuk indeks dengan menggunakan asas yang diberi dalam tanda kurung.
Write each number in index form using the base given in brackets.  SP1.1.2  TP2

16 (Asas 2/Base 2) Kaedah Pembahagian Berulang Kaedah Pendaraban Berulang
Maka/Thus, 16 = 24 Repeated Division Method Repeated Multiplication Method

n=4 2 16 2×2×2×2
2 8 4
2 4
2 2 8
1
16

1. 64 (Asas 4/Base 4) 2. 243 (Asas 3/Base 3)

4 64 4×4×4 3 243 3×3×3×3×3
n = 3 4 16 16 3 81 9
64 n = 5 3 27 27
44 39 81
1 33

1 243

Maka/Thus, 64 = 43 Maka/Thus, 243 = 35

1 SP 1.1.1   TP 1 2 3 4 5 6 SP 1.1.2   TP 1 23456
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  1.1  Tatatanda Indeks Buku Teks: m.s. 5 – 6

A. Hitung nilai bagi nombor yang berikut tanpa menggunakan kalkulator.
Calculate the values of the following numbers without using a calculator.  SP1.1.2  TP2

34 = 81 1. 44 = 256 2. 0.23 = 0.008

3×3×3×3 4×4×4×4 0.2 × 0.2 × 0.2
9 × 3 16 × 4 0.04 × 0.2
27 × 3 64 × 4 0.008
81 256

3. 25 = 32 4. (–0.6)3 = –0.216 1 2 5. – 72 2 = 4
49
2×2×2×2×2 (–0.6) × (–0.6) × (–0.6)
4 × 2 0.36 × (–0.6) 1– 72 2 × 1– 72 2
8 × 2 –0.216
16 × 2 4
32 49

B. Hitung nilai bagi nombor yang berikut dengan menggunakan kalkulator.
Calculate the values of the following numbers by using a calculator.  SP1.1.2  TP2

(a) 64 = 1 296 1. 73 = 343 2. (–9)4 = 6 561 3. 0.43 = 0.064

Tekan/Press: 4 (–1.2)3 = –1.728 1 2 5. – 43 3 = – 6247
1 2 6. 1 21 4 = 5 116
6∧4=

(b) (–2)3 = –8

Tekan/Press:

(–) 2 ∧ 3 =

C. Tentukan nilai p dalam setiap yang berikut.
Determine the value of p in each of the following.  SP1.1.2  TP2

p 3 = 27 1. p3 = –8
33 = 3 × 3 × 3 = 27 (–2)3 = (–2) × (–2) × (–2) = –8
Maka/Thus, p = 3 Maka/Thus, p = –2

2. 4 p = 256 3. – 62 = p

44 = 4 × 4 × 4 × 4 = 256 – 62 = – (6 × 6) = –36

Maka/Thus, p = 4 Maka/Thus, p = –36

2 SP 1.1.2   TP 1 2 3 4 5 6

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 1.2 Hukum Indeks Buku Teks: m.s. 6 – 8

A. Ringkaskan setiap yang berikut. NOTA
Simplify each of the following.  SP1.2.1  TP3
am × an = am + n
(a) 22 × 23 = 22 + 3 Tambahkan indeks. (b) y5 × y3 × y = y5 + 3 + 1 ↑ ↑
= 25 Add the indices. = y9 Asas yang sama
Same base

1. 34 × 32 2. 0.25 × 0.26 3. (–7)2 × (–7)4
= 34 + 2 = 0.25 + 6 = (–7)2 + 4
= 36 = 0.211 = (–7)6

4. 1 2 23 × 1 2 24 × 1 2 2 5. m2 × m3 × m5 6. n4 × n × n3
5 5 5 = m2 + 3 + 5 = n4 × n × n3
= m10 = n4 + 1 + 3
1 2 = 2 3+4+1 = n8
5

= 1 2 28
5

B. Permudahkan setiap yang berikut.
Simplify each of the following.  SP1.2.1  TP3

Operasi untuk pekali 1. 3a3 × 5a 2. 4w2 × 3w4
Operation of the coefficients = (3 × 5)a3 + 1 = (4 × 3)w2 + 4
= 15a4 = 12w6

2p2 × 4p3 = (2 × 4)(p2 × p3) 3. 1 y2 × 3y2 × 12y6 4. –6m2 × m3 × 1 m4
= 8p2 + 3 4 3
= 8p5 1 1
1 2 4 1 2 3
= × 3 × 12 y2 + 2 + 6 = –6 × m2 + 3 + 4

= 9y10 = –2m9

C. Permudahkan setiap yang berikut.
Simplify each of the following.  SP1.2.1  TP3

Kumpulkan sebutan dengan asas yang sama. 1. p3 × r × p4 × r5 2. 32 × 53 × 35 × 56
Group the terms with the same base. = (p3 × p4) × (r1 × r5) = (32 × 35) × (53 × 56)
= p3 + 4 × r1 + 5 = 32 + 5 × 53 + 6
m4 × n8 × m × n3 = p7r6 = 37 × 59
= (m4 × m1) × (n8 × n3) 3. 2a3 × b2 × 5a7 × b4
= m4 + 1 × n8 + 3 = (2 × 5)a3 × a7 × b2 × b4 4. – 4k3 × n4 × 112k2
= m5 × n11 = 10a3 + 7b2 + 4 1 2 1
= m5 n11 = 10a10b6 = – 4 × 12 (k3 × k2) × n4

Tambahkan indeks bagi sebutan dengan asas = – 13 k3 + 2 × n4
yang sama. = – 31 k5n4
Add the indices for terms with the same base.

3 SP 1.2.1   TP 1 2 3 4 5 6

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 1.2 Hukum Indeks Buku Teks: m.s. 8 – 10

A. Permudahkan setiap yang berikut.
Simplify each of the following.  SP1.2.2  TP3

NOTA

(a) 27 ÷ 23 Tolak indeks. (b) x5 = x5 ÷ x1 • am ÷ an = am – n
= 27 – 3 Subtract the indices. x   ↑    ↑
= 24 = x5 – 1 Asas yang sama
Same base
x = x 1 = x4
• am = am ÷ an
an

1. 58 ÷ 52 2. m11 ÷ m9 3. 37 ÷ 32 ÷ 34
= 58 – 2 = m11 – 9 = 37 – 2 – 4
= 56 = m2 = 31
=3

4. 76 = 76 ÷ 74 5. p6 = p6 ÷ p1 6. r8 ÷ r ÷ r2
74 p = r8 ÷ r1 ÷ r2
= 76 – 4 = r8 – 1 – 2
= 72 = p6 – 1 = r5
= p5

B. Permudahkan setiap yang berikut.
Simplify each of the following.  SP1.2.2  TP3

1 (a) 2p8 ÷ 10p2 = 2 (p8 ÷ p2) (b) x3y7 = x3y7 ÷ x1y4
10 xy4
= x3 – 1y7 – 4 Tolak indeks bagi sebutan dengan
= 1 p8 – 2 Operasi untuk pekali = x2y3 asas yang sama.
5 Operation of the coefficients Subtract the indices for terms with
the same base.
= 1 p6
5

1. 3a5 ÷ 6a2 2. 18b5 ÷ 3b4 3. 20p7q2 ÷ 4pq
= 20p7q2 ÷ 4p1q1
1  = 3 (a5 – 2) =1  18 (b5 ÷ b4)
6 3 1  20 p7 – 1q2 – 1
1 = 4
= 2 a3 = 6b5 – 4
= 6b1 = 5p6q1
= 5p6q
= 6b

1  4. 4k10 = 4 k10 – 5 1  5. 21h8 = 21 h8 – 2 6. x8y6 = x8y6 ÷ x4y1
12k5 12 7h2 7 x4y

= 1 k5 = 3h8 – 2 = x8 – 4y6 –1
3 = 3h6
= x4y5

4 SP 1.2.2   TP 1 2 3 4 5 6

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 1.2 Hukum Indeks Buku Teks: m.s. 10 – 13

A. Permudahkan setiap yang berikut.
Simplify each of the following  SP1.2.3  TP3

(a) (23)4 = 23(4) Darabkan indeks. (b) (x5)2 = x5(2) NOTA
= 212 Multiply the indices. = x10
(am)n = (an)m = amn

1. (34)4 = 34(4) 2. (73)5 = 73(5) 3. (52)6 = 52(6)
= 316 = 715 = 512

4. (h5)2 = h5(2) 5. (k3)8 = k3(8) 6. [(–p)2]4 = (–p)2(4)
= h10 = k24 = (–p)8


B. Tentukan sama ada setiap persamaan yang berikut benar atau palsu.
Determine whether each of the following equations is true or false.  SP1.2.3  TP3

(23)8 = (82)4 Sama 1. (32)6 = (92)3 2. (76)2 = (493)3
(23)8 = 224 Equal (32)6 = 312 (76)2 = 712
(82)4 = (23)2(4) = 224 (92)3 = (32)2(3) = 312 (493)3 = (72)3(3) = 718

Benar/True Benar/True Palsu/False

C. Permudahkan setiap yang berikut.
Simplify each of the following.  SP1.2.3  TP3

NOTA

(a) (23 × 74)3 (b) (2a3b)4 1 (c) 3x 2 = 32x1(2) • (ambn)q = amqbnq
= 23(3) × 74(3) = 24a3(4)b1(4) 5y5 52y5(2)
= 29 × 712 = 16a12b4 1 • am q = amq
bn bnq
9x2
= 25y10

1. (33 × 57)2 2. (p2q3r5)3 3. (7a4b3)2 4. (3h2k)3
= 33(2) × 57(2) = p2(3)q3(3)r5(3) = 72a4(2)b3(2) = 33h2(3)k1(3)
= 36 × 514 = p6q9r15 = 49a8b6 = 27h6k3

1  5. 22 5 22(5) 1  6. x2 3 x2(3) 1  7. 3c4 2 = 32c4(2) 1  8. 2m4 3 23m4(3)
73 73(5) y4 y4(3) 4d 42d1(2) 5n2 53n2(3)
= = =

= 210 = x6 = 9c8 = 8m12
715 y12 16d2 125n6

5 SP 1.2.3   TP 1 2 3 4 5 6

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 1.2 Hukum Indeks Buku Teks: m.s. 14 – 16

A. Permudahkan setiap yang berikut. NOTA
Simplify each of the following.  SP1.2.4  TP3
Hukum Indeks:
(a) 24 ÷ 24 = 24 – 4 (b) y5 ÷ y5 = y5 – 5 Laws of Indices:
= 20 = y0 • am × an = am + n
• am ÷ an = am – n
• (am)n = amn
= 1 a0 = 1 = 1 a0 = 1

1. 56 ÷ 56 = 56 – 6 2. 107 = 107 – 7 3. n8 ÷ n8 = n8 – 8 4. x15 = x15 – 15
= 50 107 = n0 x15
= 100 =1 = x0
=1
= 1 =1

B. Lengkapkan jadual yang berikut.
Complete the following table.  SP1.2.4  TP2

Bentuk indeks positif 32 x1 NOTA
Positive index form 1 1  1 1 108 2
34 p5 7 y • a–n = 1  ; a 0
x3 an ≠

Bentuk indeks negatif 3– 4 x–3 1 2p–5 7 –2 y –1 • a n = b –n
10–8 3 x b a
1  1  1  1 Negative index form

C. Permudahkan setiap yang berikut.
Simplify each of the following. SP1.2.4  TP3

(a) 32 × 35 ÷ 37 = 32 + 5 – 7 1. 44 × 47 = 44 + 7 – 11
411
= 40
= 30
=1
= 1 a0 = 1

a –2 b2
1 2 1 2 2. (a2b3)3 ×b a
a –1 = a4(2)b1(2) × b3 1 = a2(3)b3(3) ×
1  1 (b) (a4b)2 ×b3 a 1 a –n 1 b n
b = a b2
a2
= a8b2 × b3 = a6b9 ×
a
= a6 – 2b9 + 2
= a8 – 1b2 + 3 = a4b11
= a7b5
(p3q2)2 × 4p3q p3(2)q2(2) × 4p3q
(2p2q)5 25p2(5)q5
5p3 × (3p2q)2 5p3 × 32p2(2)q1(2) 3. =
15q4 15q4
(c) = = p6q4 × 4p3q
32p10q5
= 5p3 × 9p4q2 4
15q4 1 =32 p6 + 3 – 10q4 + 1 – 5

1 =5×9 p3 + 4q2 – 4 1  1 p–1q0
15 = 8

= 3p7q–2 = 1 × 1 ×1
8 p
= 3qp27  a–n = 1 1
an = 8p

6 SP 1.2.4   TP 1 2 3 4 5 6

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 1.2 Hukum Indeks Buku Teks: m.s. 17 – 20

1

A. Nyatakan setiap yang berikut dalam bentuk a n atau n a .

1 or n a .  SP1.2.5  TP2 Video

State each of the following in the form a n

na 2 36 4  81 3  1 000 3 –8 6m 5k NOTA

1 11 11 1 1 1
a n 362 814 1 0003 (–8)3
m6 k5 • n a = an Video

B. Hitung setiap yang berikut.
Calculate each of the following.  SP1.2.5  TP2

1 1 1 1 1

(a) 3 64 = 643 n a = an 1. 3 27 = 273 2. 5 32 = 325 3. 4 81 = 814
33 1 25 1 34 1
43 1  = 3  = 5  = 4 
3
= =3 =2 =3
= 41

=4 1 (–3)3 1  1 72 1  1 (–5)3 1 
3 2 3
1 (–2)3 1  4. (–27) 3 = 5. 492 = 6. (–125) 3 =
= 3 = (–3)1 = 71 = (–5)1
(b) (–8) 3
= (–2)1
= –3 = 7 = –5
= –2

C. Lengkapkan jadual yang berikut.
Complete the following table.  SP1.2.5  TP2

m am 1 a 1 m n am 1n a 2m NOTA
n n 13 8 22
an 15 y 24

1 2 1 12 3 82 m = n am = 1n am

83 (82) 3 83 • a n

1 4 1 1 4 y5 5 y4 1 m 1 1 m
y5 (y4) 5
• a n = (am)n = an


D. Hitung nilai bagi setiap yang berikut.

Calculate the value of each of the following.  SP1.2.5  TP2

2 3 Atau/Or 2 Atau/Or

325 1. 92 2. 83

Kaedah 1/Method 1: 3 = 12  9 23 3 = 32 3  2 = 13  8 22 2 = 23 2 
2 3
92 92 83 83
2 = 33 = 22
= 15 3222 = 33 = 22
325
= 22 = 27 = 27 =4 =4

= 4 2 3

Kaedah 2/Method 2: 3. 1253 Atau/Or 4. 164 Atau/Or

32 2 = 25 2  2 = 13  12522 2 = 53 2  3 = 14  1623 3 = 24 3 
5 5 3 4
125 3 125 3 164 164
= 22 = 52 = 23
= 52 = 23
=4
= 25 = 25 =8 =8

7 SP 1.2.5   TP 1 2 3 4 5 6

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 1.2 Hukum Indeks Buku Teks: m.s. 20 – 21

A. Permudahkan setiap yang berikut.

Simplify each of the following.  SP1.2.6  TP3

(m3n2)2 ÷ (m–1n)5 1  1. (a–1b3)2 × a –1
b2
b2
= m3(2)n2(2) ÷ m–1(5)n5 = a–1(2)b3(2) × a
= m6n4 ÷ m–5n5
= m6 – (–5)n4 – 5 a–2b6 b2
= × a

= m11n–1 = a–2 – 1 b6 + 2

= m11 n–1 = 1 = a–3b8
n n
b8
= a3

2. (a–2)3 × b = a–2(3) × b1 3. (3mn–1)3 × 1 m–1n2
(ab4)2 a2b4(2) 3
a–6 × b1 1
= a2b8 = 33m3n(–1)3 × 3 m–1n2

= a–6 – 2b1 – 8 = 27m3n–3 × 1 m–1n2
= a–8b–7 3
1
1 1  = 3 m3 + (–1)n–3 + 2
= a8b7 27 ×

= 9m2n–1

= 9m2
n

B. Hitung nilai bagi setiap yang berikut tanpa menggunakan kalkulator.

Calculate the value of each of the following without using a calculator  SP1.2.6  TP3

3 1. 5 × 216– 34 ÷ 1
16 4 × (24)2 ÷ 82 6
36 2

= 24 3  × 24(2) ÷ 23(2) Tukar semua nombor indeks kepada = 62 5  × 63– 43  ÷ 6–1
4 asas yang sama, iaitu asas 2. 2
Change all the index numbers to the
= 23 × 28 ÷ 26  same base, that is base 2. = 65 × 6–4 ÷ 6–1
= 65 + (–4) – (–1)
= 23 + 8 – 6
= 62
= 25
= 36
= 32

5 2 1 4
2. 27 3 × (34)2 × 9–5 52 625
5 3. 1253 × ÷
3
= 33  × 34(2) × 32(–5) = 53 2  5–2 54 1 
3 4
= 35 × 38 × 3–10 × ÷
= 52 × 5–2 ÷ 51
= 35 + 8 (–10) = 52 + (–2) –1

= 33 = 5–1

= 27 = 1
5

8 SP 1.2.6   TP 1 2 3 4 5 6

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  1.2  Hukum Indeks Buku Teks: m.s. 20 – 21

A. Permudahkan setiap yang berikut.
Simplify each of the following.  SP1.2.6  TP3

1 (6) 1
2
(m2n)3 × 1 mn6 = (m2)3n3 × (mn) 1. (81x6 y8) 4 ÷ xy2

1 m3 1 n6 1  = 1 x6 1 y8 1  ÷ 1
3 3 4 4
(m3n6) 3 81 4 (xy2) 2

= m6n3 × (mn)3 = 34 1 x 3 y2 ÷ x 1 y2 1 
m1n2 4 2 2 2

m6n3 × m3n3 31
m1n2
= = 3x 2 y2 ÷ x 2 y1

= m6 + 3 – 1n3 + 3 – 2 = 3 – 1 – 1
= m8n4
3x 2 2 y2

= 3x1y1

= 3xy

1 1 1 3 3. (–2a)3 × (3a3b–2)2
72a4b5
2. 2m 2 n 6 × m5n–3

= 23m 1 (3)n 1 (3) × 1 = (–2)3a3 × 32a3(2)b–2(2)
2 6 72a4b5
(m5n–3) 2

= 31 × m5 1 n–3 1  = –8a3 × 9a6b– 4
2 2 72a4b5
8m 2 n 2

= 31 × m 5 n– 32 1  = –8 × 9 a3 + 6 – 4b–4 – 5
2 72
8m 2 n 2

= 3 + 51 +– 32  = (–1)a5b–9

8m 2 2n2
= 8m4n–1
8m4 = – ab95
= n

B. Hitung setiap yang berikut.

Calculate each of the following.  SP1.2.6  TP3

1  1. 2 32 2 (2) 3 (2) 3
5 2
3 27– 31 32 5 × 52 = 32 × 5 2. (23 × 32 × 5)2
× × 254 32 × 3 27 × 5
16 4 4 16 1 254 1 
2
1 16 4 × 3 3 3
23 2  32 2 
[24 × 32] 2 4 = × × 52

24 3  × 33– 31  = 32 5 × 53 1 1 1
4
= 1 32 2 × 27 3 × 5 2

24 1  × 32 1  16 4 × 252 93
2 2 4
25 5  22 33 × 52
= 23 × 3–1 × 53 = ×
22 × 31 = 25 1  × 33 1  1
24 1  2 3
4 52(2) × 52
= 23 – 2 × 3–1 – 1 × 93
= 21 × 3–2 24 53
= 21 × 54 = 22 × 33 × 52
×
= 2 × 1 = 24 – 1 × 53 – 4 5 1
9
= 23 × 5–1 2 2 × 31 × 5 2

= 2 1 = 9 – 5 × 33 – 1 × 3 – 1
9 5 2 2
= 8 × 22 52
= 22 × 32 × 51
= 8 = 180
5

9 SP 1.2.6   TP 1 2 3 4 5 6

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  1.2  Hukum Indeks Buku Teks: m.s. 22 – 23

A. Cari nilai x bagi setiap persamaan yang berikut.
Find the value of x for each of the following equations.  SP1.2.7  TP4

2x × 4x + 3 = 1 1. 26 ÷ 2x = 16x – 1

2x × 4x + 3 = 1 1 = 20 26 ÷ 2x = 16x – 1
2x × 22(x + 3) = 20 26 – x = 24(x – 1)
26 – x = 24x – 4
2x + 2(x + 3) = 20
2x + 2x + 6 = 20 Maka/Thus, 6 – x = 4x – 4
23x + 6 = 20
6 + 4 = 4x + x

Maka/Thus, 3x + 6 = 0  Samakan indeks. 10 = 5x
3x = –6 Equate the indices.
x = –2 x=2

2. 3x × 27x – 2 = 9 3. 81x ÷ 3x – 1 = 1
9
3x × 27x – 2 = 9
3x × 33(x – 2) = 32 34(x) ÷ 3x – 1 = 1
3x + 3(x – 2) = 32 32
3x + 3x – 6 = 32 34x ÷ 3x – 1 = 3–2
34x – 6 = 32 34x – (x – 1) = 3–2
33x + 1 = 3–2
Maka/Thus, 4x – 6 = 2
4x = 8 Maka/Thus, 3x + 1 = –2
x=2
3x = –3

x = –1

B. Cari nilai-nilai x yang mungkin bagi setiap persamaan yang berikut.
Find the possible values of x for each of the following equations.  SP1.2.7  TP4

1. 3x2 = 96 – 2x HEBAT MODUL 31 2. 2x2 – 3x = 16

3x2 = 32(6 – 2x) 2x2 – 3x = 24
3x2 = 312 – 4x

x2 = 12 – 4x x2 – 3x = 4
x2 – 3x – 4 = 0
x2 + 4x – 12 = 0  Persamaan kuadratik (x + 1)(x – 4) = 0
(x + 6)(x – 2) = 0 A quadratic equation

x + 6 = 0 atau/or x – 2 = 0 x + 1 = 0 atau/or x – 4 = 0

x = –6 x=2 x = –1 x=4

Nilai-nilai x yang mungkin ialah –6 dan 2. Nilai-nilai x yang mungkin ialah –1 dan 4.
The possible values of x are –6 and 2. The possible values of x are –1 and 4.

10 SP 1.2.6   TP 1 2 3 4 5 6

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 1.2 Hukum Indeks Buku Teks: m.s. 22 – 23

A. Hitung nilai bagi setiap yang berikut tanpa menggunakan kalkulator.
Calculate the value of each of the following without using a calculator.  SP1.2.7  TP4

1. 5 × 3 × 4– 14 2. 3 × (4 × 9)– 21 ÷ 2

10 2 81 4 83

= 1 × (5 × 3 × 22– 14  = 34 3  × (4 × 9)– 12 ÷ 23 2 
4 3
52 2) 2

= 1 × 3 × 3 × 2– 12 = 33 × 22– 12  × 32– 12  ÷ 22
= 33 × 2–1 × 3–1 ÷ 22
52 52 22 = 33 + (–1) × 2–1 – 2
= 32 × 2–3
= 1 + 3 × 3 – 1
2 2
52 22

= 52 × 21 = 9× 1
= 50 8

= 9
8

B. Selesaikan persamaan serentak yang berikut.
Solve the following simultaneous equations.  SP1.2.7  TP4

1. 5x × 252y = 5 dan/and 3x × 9y – 1 = 1 2. 9x × 3y = 37 dan/and 2x ÷ 2y = 1
27 2

5x × 252y = 5 ……① 9x × 3y = 37 ……① Kaedah penggantian:
5x × 52(2y) = 51 32(x) × 3y = 37 ……② Substitution method:
5x + 54y = 51 32x + 3y = 37
x + 4y = 1 2x + y = 7 Dari/From:
2x + y = 7……①
3x × 9y – 1 = 1 2x ÷ 2y = 1 y = 7 – 2x……③
27 2
2x – y = 2–1
3x × 32(y – 1) = 1
33 x – y = –1

3x × 32y – 2 = 3–3 ① + ②: 3x = 6 Gantikan ③ ke dalam ②.
3x + 2y – 2 = 3–3 x=2 Substitute ③ into ②.
x – y = –1
x + 2y – 2 = –3 Gantikan x = 2 ke dalam ①. x – (7 – 2x) = –1
Substitute x = 2 into ①. x – 7 + 2x = –1
x + 2y = –1 ……② 3x = 6
x = 2
① – ②: 2y = 2
y=1 2(2) + y = 7 Gantikan x = 2 ke dalam
③.
Gantikan y = 1 ke dalam ①. 4+y=7 Substitute x = 2 into ③.
Substitute y = 1 into ①. y = 7 – 2(2)
y=3 =3
Maka/Thus,
x + 4(1) = 1 Maka/Thus, x = 2, y = 3 x = 2, y = 3
x+4=3
x = –1

Maka/Thus, x = –1, y = 1

11 SP 1.2.7   TP 1 2 3 4 5 6

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  1.2  Hukum Indeks Buku Teks: m.s. 24

Selesaikan masalah yang berikut.
Solve the following problems.  SP1.2.7  TP5

1. Jeya membeli sebuah rumah dengan harga 2. Lima tahun yang lepas, Encik Lee membeli sebuah
RM210 000. Purata kadar inflasi dalam tempoh
10 tahun ialah 3% dikompaun setiap tahun. kereta baharu dengan harga RM57 000. Kini dia
Jeya bought a house for RM210 000. The average
inflation rate over a 10 year period was 3% ingin menjual kereta itu kepada peniaga kereta
compounded annually.
terpakai. Harga kereta itu akan dihitung dengan
(a) Tulis satu ungkapan bagi harga rumah, dalam
RM, selepas t tahun. 1 2formula 8 n, dengan ialah bilangan
Write an expression for the price of the house, in RM57   000  9 n
RM, after t years.
tahun selepas sebuah kereta dibeli.
(b) Hitung harga rumah itu selepas 7 tahun.
Calculate the price of the house after 7 years. Five years ago, Mr Lee bought a new car for RM57   000.

Mengaplikasi Now he intends to sell the car to a used car dealer. The

price of the car will be calculated using the formula
8
1 2RM57   000 9 n  where n is the number of years after

,

a car is bought.

(a) Harga rumah selepas (a) Hitung nilai pasaran kereta Encik Lee. Beri
Price of house after: jawapan kepada RM yang terdekat.
Calculate the market value of Mr Lee’s car. Give
1 tahun/year: the answer to the nearest RM.
RM210 000 + 3% × RM210 000
= RM210 000(1 + 0.03) (b) Berapakah kerugian Encik Lee dengan menjual
= RM210 000(1.03) kereta itu?
How much is Mr Lee’s loss by selling the car?
2 tahun/years:
RM210 000(1.03)(1.03) Mengaplikasi
= RM210 000(1.03)2
(a) n = 5
3 tahun/years:
RM210 000(1.03)2 (1.03) Nilai pasaran kereta Encik Lee
= RM210 000(1.03)3
Market value of Mr Lee’s car

1 2 8 5
= RM57 000  9

= RM31 630.95

= RM31 631

t tahun/years: RM210 000(1.03)t (kepada RM yang terdekat)

(to the nearest RM)

(b) Apabila/When t = 7, (b) Kerugian Encik Lee

Harga rumah/Price of house Mr Lee’s loss
= RM210 000(1.03)7 = RM57 000 – RM31 631
= RM258 273.51 = RM25 369

12 SP 1.2.7   TP 1 2 3 4 5 6

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PRAKTIS PT3

  Bahagian A   Bahagian B
8. (a) Padankan pasangan indeks yang setara.
1. 2
KLON 3 5 = b  ac Match the equivalent pairs of indices.
PT3 [3 markah/3 marks]

Cari nilai-nilai a, b dan c. p p–3

Find the values of a, b and c. 1 5 p3
p3
A a = 2, b = 3, c = 5

B a = 5, b = 2, c = 3

C a = 3, b = 2, c = 5

D a = 3, b = 5, c = 2


2. Diberi 2m × 27 = 221, nilai m ialah 31
Given that 2m × 27 = 221, the value of m is
A 3 p5 p2
B 12
C 14 (b) Nyatakan sama ada Benar atau Palsu.
D 28 State whether True or False.
[1 markah/1 mark]
4
B 4 p5 6–2 = 1 Palsu/False
3. p5 = D 1 p4 25 12

A 5 p4
C 1 p5 24

4. Antara berikut, yang manakah tidak sama dengan 9. (a) Lengkapkan rajah di bawah dengan nilai
yang betul.
64? Complete the following diagram with the correct
values.
Which of the following is not equal to 64? [3 markah/3 marks]

HEBAT MODUL 31   a11 ÷ a 5  

A 43 B 82
C 64 D 26

5. Diberi (yn)3 = y27, nilai n ialah

Given that (yn)3 = y27, the value of n is

A 3 B 9 a6

C 12 D 24 a2 × a 4  a 4 3
2

6. Antara berikut, yang manakah setara dengan 3– 4? (b) 2
Which of the following is equivalent to 3–4?
643 = 16

A 1 B – 413
45
1 D – 314 Tulis Benar atau Palsu.
C 34 Write True or False.

Benar/True [1 markah/1 mark]

7. 32 × 3n × 273 = B 318n
A 36n D 3n + 11
C 3n + 5

13

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  Bahagian C (ii) Hitung jisim bahan yang tinggal selepas
6 jam.
10. (a) Permudahkan: Calculate the remaining mass after 6 hours.
[2 markah/2 marks]
KLON Simplify:
PT3 Mengaplikasi

(i) q5 ÷ q2
1
(ii) Apabila/When t = 6,
(ii) (9y2w6)2
1  1 6
[4 markah/4 marks] M = 100 g × 2 = 1.5625 g

(i) q5 ÷ q2 = q3

1 1 y2 1 w6 1  11. (a) Ungkapkan setiap yang berikut dalam bentuk
2 2 2 7n.
(ii) (9y2w6) 2 = 9
= 3yw3 Express each of the following in the form 7n.

(b) Gunakan jadual di bawah untuk mengungkap (i) 1 (ii) 74 × 4910
nilai bagi 256 × 4 096 sebagai nombor kuasa 4. 7 [3 markah/3 marks]
Use the table below to express the value of
256 × 4 096 as a number to the power of 4. (i) 7– 12 (ii) 74 × 4910 = 74 × 72(10)
= 74 + 20

41 42 43 44 45 46 = 724
4 16 64 256 1 024 4 096
(b) (i) Diberi 81 – x = 4, cari nilai x.

Given that 81 – x = 4, find the value of x.
[2 markah/2 marks]
[2 markah/2 marks]
TIMSS

256 × 4 096 = 44 × 46 23(1 – x) = 22
23 – 3x = 22
= 44 + 6
= 410 3 – 3x = 2

–3x = –1
1
(c) Satu sampel bahan radioaktif dengan x = 3
jisim
awal 100 g mereput dari masa ke
masa, menjadi separuh setiap jam. (ii) Tunjukkan bahawa: 9– 12 × 1 = 3– 4
A sample of radioactive substance with an initial Show that: 27
mass of 100 g decays over time, halving every
HEBAT MODUL 31
hour.
(i) Cari satu rumus bagi jisim, M g, bahan [2 markah/2 marks]

yang tinggal selepas t jam. 9– 12 × 1 = 32– 12  × 1
Find a formula for the remaining mass, 27 33
M g, after t hours. = 3–1 × 3–3
= 3–4
[2 markah/2 marks]

Selepas/After
(c) Diberi 1 3 2n = 237xy, ungkapkan n dalam sebutan
1 3 2n = n in terms of x
1 jam/hour: M = 100 g × 1 KLON x dan y. 3x
2 PT3 27y

1  12 Given that , express
2
2 jam/hours: M = 100 g × and y.

13 [3 markah/3 marks]
2
1 
3 jam/hours: M = 100 g × 3 1 (n) = 3x
2 = 33(y)
3x – 3y
1  1t 1
t jam/hours: M = 100 g × 2 3 2 n

1 n = x – 3y
2
n = 2(x – 3y)

atau/or n = 2x – 6y

14

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