Indices, Surds and Logarithms: Additional Mathematics questions

Section A: objective questions

14 marks

Choose the correct answer. Each question carries 1 mark.

Find the value of 8²³.

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8^2/3 = (³√8)² = 2² = 4.

For the teacher: Understanding, low difficulty, standard 4.1.2

Textbook p. 92

Simplify (x³y⁻²)²x³y⁻³.

x³y⁻⁷
x³y
x³y⁻¹
x⁶y⁻¹

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Numerator x⁶y⁻⁴. Subtract the indices: x⁶⁻³y^{−4−(−3)} = x³y⁻¹.

For the teacher: Applying, medium difficulty, standard 4.1.1

Textbook p. 90

Solve 3^x+1 = 27.

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27 = 3³, so x + 1 = 3 and x = 2. (x = 3 is the index of 27 itself, forgetting the +1.)

For the teacher: Applying, low difficulty, standard 4.1.3

Textbook p. 93

Find x if 2^x+2 − 2^x = 24.

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Factorise: 2^x(4 − 1) = 24, so 2^x = 8 and x = 3.

For the teacher: Analysing, high difficulty, standard 4.1.3

Textbook p. 93

Which of the following is an irrational number?

√16
√12
√94
0.25

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√16 = 4, √(9/4) = 3/2 and 0.25 = 1/4 are rational. √12 cannot be written as a fraction of integers.

For the teacher: Remembering, low difficulty, standard 4.2.1

Textbook p. 96

Simplify √48 + √12.

√60
5√3
6√3
8√3

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√48 = 4√3 and √12 = 2√3, so the sum is 6√3. √60 wrongly adds the numbers under the roots.

For the teacher: Applying, low difficulty, standard 4.2.2

Textbook p. 99

Rationalise the denominator of 6√3.

63
6√3
3√2
2√3

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Multiply top and bottom by √3: 6√3 ÷ 3 = 2√3.

For the teacher: Applying, low difficulty, standard 4.2.3

Textbook p. 103

Express 13 + √5 with a rational denominator.

3 − √54
3 + √54
3 − √514
3 − √52

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Multiply by the conjugate: (3 − √5) ÷ (9 − 5) = (3 − √5)/4.

For the teacher: Applying, medium difficulty, standard 4.2.3

Textbook p. 103

Which is the value of log₂ 32?

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2⁵ = 32, so the logarithm is 5.

For the teacher: Remembering, low difficulty, standard 4.3.1

Textbook p. 108

Evaluate log₃ 6 + log₃ 1.5.

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Product law: log₃(6 × 1.5) = log₃ 9 = 2.

For the teacher: Applying, low difficulty, standard 4.3.2

Textbook p. 113

Given log_a 2 = p and log_a 3 = q, express log_a 12 in terms of p and q.

p + 2q
2p × q
4p + q
2p + q

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12 = 2² × 3, so log_a 12 = 2p + q.

For the teacher: Applying, low difficulty, standard 4.3.2

Textbook p. 114

Which expression is equal to log₄ 8?

log₂ 8log₂ 4
log₂ 4log₂ 8
log₂ 8 − log₂ 4
log₂ 2

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Change of base to base 2: log_a b = log_c b ÷ log_c a, which equals 3/2.

For the teacher: Understanding, medium difficulty, standard 4.3.3

Textbook p. 116

Given 3^x = 20, find x correct to 2 decimal places.

0.37
2.73
6.67
1.82

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x = lg 20 ÷ lg 3 = 1.3010 ÷ 0.4771 = 2.73. 6.67 is 20 ÷ 3.

For the teacher: Analysing, medium difficulty, standard 4.3.4

Textbook p. 118

Solve log₂(x + 1) = 3.

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Index form: x + 1 = 2³ = 8, so x = 7. (8 forgets to subtract 1.)

For the teacher: Analysing, low difficulty, standard 4.3.4

Textbook p. 119

Section B: structured questions

31 marks

Answer all questions. Show your working.

Find the value of each expression without a calculator.

[1 mark] (a)5³ × 5⁻¹ ÷ 5²

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5³⁻¹⁻² = 5⁰ = 1 [1]1

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[1 mark] (b)27²³

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(³√27)² = 3² = 9 [1]1

Tick each point you got.

[2 marks] (c)(1681)³⁴

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Fourth root: 2/3 [1]1
(2/3)³ = 8/27 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 4.1.2

Textbook p. 92

Solve each equation.

[2 marks] (a)2^3x−1 = 32

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32 = 2⁵ [1]1
3x − 1 = 5, x = 2 [1]1

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[2 marks] (b)9^x = 27^x−1

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3^2x = 3^3x−3 [1]1
2x = 3x − 3, x = 3 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 4.1.3

Textbook p. 93

Simplify, leaving the answers in surd form.

[2 marks] (a)√75 + √12 − √27

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5√3 + 2√3 − 3√3 [1]1
4√3 [1]1

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[2 marks] (b)(2 + √3)²

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4 + 4√3 + 3 [1]1
7 + 4√3 [1]1

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[1 mark] (c)(3 + √2)(3 − √2)

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9 − 2 = 7 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 4.2.2

Textbook p. 99

Rationalise the denominator and simplify.

[1 mark] (a)10√5

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10√5 ÷ 5 = 2√5 [1]1

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[2 marks] (b)√3√3 − 1

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Multiply by (√3 + 1)/(√3 + 1): (3 + √3)/(3 − 1) [1]1
(3 + √3)/2 [1]1

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[2 marks] (c)Express 2 + √32 − √3 in the form p + q√3.

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(2 + √3)² ÷ (4 − 3) [1]1
7 + 4√3, p = 7, q = 4 [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 4.2.3

Textbook p. 103

Given log₅ 2 = m and log₅ 3 = n, express in terms of m and n.

[1 mark] (a)log₅ 6

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m + n [1]1

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[1 mark] (b)log₅ 12

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2m + n [1]1

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[2 marks] (c)log₅ 185

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log₅ 18 − log₅ 5 [1]1
m + 2n − 1 [1]1

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For the teacher: Applying, low difficulty, standard 4.3.2

Textbook p. 113

Solve the equations.

[3 marks] (a)log₂(x + 3) + log₂ x = 2

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x(x + 3) = 2² = 4 [1]1
x² + 3x − 4 = 0, so x = 1 or x = −4 [1]1
Reject x = −4 because log x is undefined; x = 1 [1]1

Tick each point you got.

[2 marks] (b)4^x = 7, correct to 3 significant figures

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x = lg 7 ÷ lg 4 = 0.8451 ÷ 0.6021 [1]1
x = 1.40 [1]1

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For the teacher: Analysing, high difficulty, standard 4.3.4

Textbook p. 118

Siti puts RM5 000 in an account that pays 5% compound interest a year. The amount after n years is 5 000(1.05)ⁿ.

[3 marks] (a)Find the least whole number of years for her money to exceed RM6 500.

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1.05ⁿ > 1.3 [1]1
n > lg 1.3 ÷ lg 1.05 = 0.1139 ÷ 0.02119 = 5.38 [1]1
n = 6 years [1]1

Tick each point you got.

[1 mark] (b)Her brother says that 5 years is enough. Is he correct? Show a calculation.

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5 000(1.05)⁵ = RM6 381.41, which is below RM6 500, so he is not correct [1]1

Tick each point you got.

For the teacher: Evaluating, medium difficulty, standard 4.4.1

Textbook p. 122

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Section B: structured questions

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21