 # Frank Solutions for Class 9 Maths Chapter 9 Indices

Frank Solutions for Class 9 Maths Chapter 9 Indices is an important study material for the students, from an exam point of view. Students can use these solutions designed by the subject matter experts who hold vast conceptual knowledge. Practising these solutions on a daily basis helps them to analyze the important questions that would appear in the final examination. Those who aim to score more marks in exams, can download Frank Solutions for Class 9 Maths Chapter 9 Indices PDF, from the links which are available below.

Chapter 9 has problems based on Indices. The solutions are curated in such a way that students grasp the concepts effortlessly. These solutions are available both online and offline, 24 / 7 as per their requirements.

## Frank Solutions for Class 9 Maths Chapter 9 Indices Download PDF                     ## Access Frank Solutions for Class 9 Maths Chapter 9 Indices

1. Evaluate the following:

(i) 60

(ii) (1 / 2)-3

(iii) (iv) (v) (0.008)(2 / 3)

(vi) (0.00243)(-3 / 5)

(vii) (viii) Solution:

(i) 60 = 1

(ii) (1 / 2)-3 = (2)3

We get,

(1 / 2)-3 = 8

(iii) = 26

We get,

= 64 (iv) = 36

We get,

= 729 (v) (0.008)(2 / 3) = (0.23)(2 / 3)

= (0.2)3 × 2 / 3

= (0.2)2

We get,

(0.008)(2 / 3) = 0.04

(vi) (0.00243)(-3 / 5) = {1 / (0.00243)(3 / 5)}

= {1 / (0.35)(3 / 5)}

= {1 / (0.3)3}

We get,

(0.00243)(-3 / 5) = (1 / 0.027)

(vii) = = = 56 × 1 / 6

We get,

= 5 (viii) = (64 / 27)(2 / 3)

= (4 / 3)3 × 2 / 3

= (4 / 3)2

We get, = (16 / 9)

2. Evaluate the following:

(a) 94 ÷ 27(-2 / 3)

(b) 7-4 × (343)(2 / 3) ÷ (49)(- 1 / 2)

(c) (64 / 216)(2 / 3) × (16 / 36)(- 3 / 2)

Solution:

(a) 94 ÷ 27(-2 / 3) = {(3)2}4 ÷ {(3)3}(-2 / 3)

On further calculation, we get,

= (3)2 × 4 ÷ (3)3 × (- 2 / 3) [Using (am)n = amn]

= (3)8 ÷ (3)-2

= (3)8 – (-2) [Using am ÷ an = am – n]

= (3)8 + 2

= (3)10

This can be written as,

= (3)2 × 5

= {(3)2}5

= (9)5

We get,

= 59049

(b) 7-4 × (343)(2 / 3) ÷ (49)(- 1 / 2)

On calculating further, we get,

= 7-4 × (73)(2 / 3) ÷ (72)(- 1 / 2)

= 7-4 × 73 × 2 / 3 ÷ 72 × (- 1 / 2)

= 7-4 × 72 ÷ 7-1

= 7-4 + 2 – (-1) [Using am × an = am + n and am ÷ an = am – n]

= 7-4 + 2 + 1

We get,

= 7-1

= (1 / 7) [Using a-m = (1 / am)]

(c) (64 / 216)(2 / 3) × (16 / 36)(- 3 / 2)

On further calculation, we get,

= {(26)2 / 3 / (63)2 / 3} × {(24)-3 / 2 / (62)– 3 / 2}

= {(2)6 × 2 / 3 / (6)3 × 2 / 3} × {(2)4 × (-3 / 2) / (6)2 × (- 3 / 2)}

[Using (am)n = amn]

= {(2)2 × 2 / (6)2} × {(2)2 × (- 3) / (6)– 3}

= {(2)4 / (6)2} × {(2)-6 / (6)-3}

We get,

= {(2)4 / (6)2} × {(6)3 / (2)6}

[Using a-m = (1 / am)

= {(2)4 / (2)6} × {(6)3 / (6)2}

= (2)4 – 6 × (6)3 – 2

[Using am ÷ an = am – n]

= (2)-2 × (6)1

= (1 / 22) × 6

= (1 / 4) × 6

We get,

= (3 / 2)

3. Write each of the following in the simplest form:

(a) (a3)5 × a4

(b) a2 × a3 ÷ a4

(c) a1 / 3 ÷ a-2 / 3

(d) a-3 × a2 × a0

(e) (b-2 – a-2) ÷ (b-1 – a-1)

Solution:

(a) (a3)5 × a4 = (a)3 × 5 × a4

[Using (am)n = amn], we get,

= (a)15 × a4

= a15 + 4

[Using am × an = am + n]

We get,

= a19

(b) a2 × a3 ÷ a4 = a2 + 3 – 4

[Using am × an = am + n and am ÷ an = am – n]

We get,

= a1

= a

(c) a1 / 3 ÷ a-2 / 3 = a(1 / 3) – (- 2 / 3)

[Using am ÷ an = am – n]

We get,

= a(1 / 3) + (2 / 3)

= a(1 + 2) / 3

= a1

= 0

(d) a-3 × a2 × a0 = a-3 + 2 + 0

[Using am × an = am + n]

= a-1

We get,

= (1 / a)

(e) (b-2 – a-2) ÷ (b-1 – a-1)

This can be written as,

= (1/b2 – 1/a2) / (1/b – 1/a)

= {(1/b)2 – (1/a)2} / (1/b – 1/a)

= {(1/b + 1/a) (1/b – 1/a)} / (1/b – 1/a)

We get,

= (1/b + 1/a)

4. Evaluate the following:

(i) (23 × 35 × 242) / (122 × 183 × 27)

(ii) (43 × 37 × 56) / (58 × 27 × 33)

(iii) (122 × 75-2 × 35 × 400) / (482 × 15-3 × 525)

(iv) (26 × 5-4 × 3-3 × 42) / (83 × 15-3 × 25-1)

Solution:

(i) (23 × 35 × 242) / (122 × 183 × 27)

This can be written as,

= {23 × 35 × (23 × 3)2} / (22 × 3)2 × (2 × 32)3 × (33)

= (23 × 35 × 26 × 32) / (24 × 32 × 23 × 36 × 33)

= (29 × 37) / (27 × 311)

On further calculation, we get,

= (29 – 7 / 311 – 7)

= 22 / 34

We get,

= (4 / 81)

(ii) (43 × 37 × 56) / (58 × 27 × 33)

This can be written as,

= (22)3 × 37 – 3} / (58 – 6 × 27)

= (26 × 37 – 3) / (58 – 6 × 27)

On further calculation, we get,

= (26 × 34) / (52 × 27)

= {34 / (52 × 27 – 6)}

= {81 / (52 × 21)}

We get,

= (81 / 50)

(iii) (122 × 75-2 × 35 × 400) / (482 × 15-3 × 525)

This can be written as,

= (122 × 35 × 400 × 153) / (482 × 525 × 752)

= {(22 × 3)2 × (7 × 5) × (24 × 52) × (3 × 5)3} / {(24 × 3)2 × (3 × 52 × 7) × (3 × 52)2}

On calculating further, we get,

= (24 + 4 × 32+ 3 × 51 + 2 + 3 × 7) / (28 × 32 + 1 + 2 × 54 + 2 × 7)

= (28 × 35 × 56 × 7) / (28 × 35 × 56 × 7)

We get,

= 1

(iv) (26 × 5-4 × 3-3 × 42) / (83 × 15-3 × 25-1)

= (26 × 42 × 153 × 251) / (83 × 54 × 33)

This can be written as,

= {(26 × (22)2 × (3 × 5)3 × (52)1} / {(23)3 × 54 × 33}

= (26 + 4 × 33 × 53 + 2) / (29 × 33 × 54)

= (210 – 9 × 55 – 4)

= (2 × 5)

We get,

= 10

5. Simplify the following and express with positive index:

(a) 3p-2q3 ÷ 2p3q-2

(b) {(p-3)2 / 3}1 / 2

Solution:

(a) 3p-2q3 ÷ 2p3q-2

This can be written as,

= (3p-2q3) / (2p3q-2)

= (3 / 2) {(p-2 / p3) × (q3 / q-2)}

= (3 / 2) {(p-2 ÷ p3) × (q3 ÷ q-2)}

= (3 / 2) {(p– 2 – 3) × (q3 – (- 2))} [Using am ÷ an = am – n]

= (3 / 2) {(p-5) × (q5)}

= (3 / 2) {(1 / p5) × (q5)}

We get,

= (3q5 / 2p5)

(b) {(p-3)2 / 3}1 / 2

= p-3 × (2 / 3) × (1 / 2)

[Using (am)n = amn]

= p– 1

We get,

= (1 / p)

6. Evaluate the following:

(i) {1 – (15 / 64)}– 1 / 2

(ii) (8 / 27)– 2 / 3 – (1 / 3)– 2 – 70

(iii) 95 / 2 – 3 × 50 – (1 / 81) – 1 / 2

(iv) (27)2 / 3 × (8)– 1 / 6 ÷ (18)– 1 / 2

(v) (16)3 / 4 + 2 (1 / 2)-1 × 30

(vi) Solution:

(i) {1 – (15 / 64)}– 1 / 2

On taking LCM, we get,

= {(64 – 15) / 64}– 1 / 2

= (49 / 64) – 1 / 2

= (64 / 49)1 / 2

We get,

= (8 / 7)

(ii) (8 / 27)– 2 / 3 – (1 / 3)– 2 – 70

= (27 / 8)2 / 3 – (1 / 3)-2 – 70

= (27 / 8)2 / 3 – (3)2 – 1

On further calculation, we get,

= (3 / 2)3 × 2 / 3 – 9 – 1

= (3 / 2)2 – 10

= (9 / 4) – 10

On taking LCM, we get,

= {(9 – 40) / 4}

= (- 31 / 4)

(iii) 95 / 2 – 3 × 50 – (1 / 81) – 1 / 2

On further calculation, we get,

= 32 × 5 / 2 – 3 × 1 – (81)1 / 2

= 35 – 3 – 92 × 1 / 2

= 243 – 3 – 9

We get,

= 231

(iv) (27)2 / 3 × (8)– 1 / 6 ÷ (18)– 1 / 2

This can be written as,

= 33 × 2 / 3 × {1 / (23 × 1 / 6)} ÷ (1 / 18)1 / 2

= (32) / (21 / 2) × (2 × 32)1 / 2

= (32 / 21 / 2) × 21 / 2 × 3

We get,

= 32 + 1

= 33

= 27

(v) (16)3 / 4 + 2 (1 / 2)-1 × 30

= 24 × 3 / 4 + 2 × 2 × 1

On further calculation, we get,

= 23 + 4

= 8 + 4

We get,

= 12

(vi) = (1 / 22)1 / 2 + (0.1)-1 – 32

= (1 / 2) + (0.1)-1 – 32

= (1 / 2) + (1 / 0.1) – 9

= (1 / 2) + (10 / 1) – 9

= (1 / 2) + 1

On taking LCM, we get,

= {(1 + 2) / 2}

= (3 / 2)

7. Simplify the following:

(a) (27x9)2 / 3

(b) (8x6y3)2 / 3

(c) (64a12 / 27b6)– 2 / 3

(d) (36m-4 / 49n-2)– 3 / 2

(e) (a1 / 3 + a– 1 / 3) (a2 / 3 – 1 + a– 2 / 3)

(f) ÷ (g) {(am){m – (1 / m)}}(1 / m + 1)

(h) xm + 2n. x3m – 8n ÷ x5m – 60

(i) (81)3 / 4 – (1 / 32)– 2 / 5 + 81 / 3. (1 / 2)-1. 20

(j) (27/343)2/3 ÷ {1/ (625/1296)1/4} x 536 / Solution:

(a) (27x9)2 / 3

This can be written as,

= (33x9)2 / 3

= (33)2 / 3(x9)2 / 3 [Using (a × b)n = an × bn]

On calculating further,

We get,

= (3)3 × 2 / 3(x)9 × 2 / 3 [Using (am)n = amn]

= (3)2x3 × 2

We get,

= 9x6

(b) (8x6y3)2 / 3

This can be written as,

= (23x6y3)2 / 3

= (23)2 / 3 (x6)2 / 3(y3)2 / 3 [Using (a × b)n = an × bn]

= (2)3 × 2 / 3(x)6 × 2 / 3(y)3 × 2 / 3 [Using (am)n = amn]

= (2)2(x)2 × 2(y)2

We get,

= 4x4y2

(c) (64a12 / 27b6)– 2 / 3

This can be written as,

= {(26a12) / (33b6)}– 2 / 3

= {26 × (- 2 / 3)a12 × (- 2 / 3)} / {33 × (- 2 / 3)b6 × (- 2 / 3)}

[Using (a × b)n = an × bn and (a / b)n = (an / bn)]

On further calculation, we get,

= (2– 4a– 8) / (3– 2b– 4)

= (32b4) / (24a8) [Using a-n = (1 / an)]

We get,

= (9b4 / 16a8)

(d) (36m-4 / 49n-2)– 3 / 2

This can be written as,

= {(62m-4) / (72n– 2)}– 3 / 2

= {62 × (- 3 / 2) m-4 × (- 3 / 2)} / {72 × (- 3 / 2) n– 2 × (- 3 / 2)}

[Using (a × b)n = an × bn and (a / b)n = (an / bn)]

On further calculation, we get,

= (6-3 m6) / (7– 3n3)

= (73m6) / (63n3)

[Using a– 1 = (1 / an)]

We get,

= (343m6) / (216n3)

(e) (a1 / 3 + a– 1 / 3) (a2 / 3 – 1 + a– 2 / 3)

= a1 / 3(a2 / 3 – 1 + a– 2 / 3) + a– 1 / 3(a2 / 3 – 1 + a– 2 / 3)

On simplification, we get,

= (a1 / 3 × a2 / 3 – a1 / 3 × 1 + a1 / 3 × a– 2 / 3) + (a– 1 / 3 × a2 / 3 – a– 1 / 3 × 1 + a– 1 / 3 × a– 2 / 3)

= {a(1 / 3) + (2 / 3) – a1 / 3 × 1 + a(1 / 3) + (- 2 / 3)} + {a(- 1 / 3) + (2 / 3) – a– 1 / 3 × 1 + a(- 1 / 3) + (- 2 / 3)}

[Using am × an = am + n]

= {a1 – a1 / 3 + a– 1 / 3} + {a1 / 3 – a– 1 / 3 + a– 1}

= {a – a1 / 3 + a– 1 / 3 + a1 / 3 – a– 1 / 3 + (1 / a)}

We get,

= {a + (1 / a)}

(f) ÷ This can be written as,

= (x4y2)1 / 3 ÷ (x5y– 5)1 / 6

On calculating further, we get,

= {x4 × (1 / 3) y2 × (1 / 3)} ÷ {x5 × (1 / 6) y– 5 × (1 / 6)}

[Using (am)n = amn]

= {x(4 / 3) y(2 / 3)} ÷ {x(5 / 6) y(- 5 / 6)}

= {x(4 / 3)y(2 / 3)} / {x(5 / 6) y(- 5 / 6)}

= x(4 / 3) – (5 / 6) y(2 / 3) – (- 5 / 6) [Using am ÷ an = am – n]

= x(1 / 2) y(3 / 2)

= x(1 / 2) (y3)(1 / 2)

[Using (am)n = amn]

= x y3

We get,

= (g) {(am){m – (1 / m)}}(1 / m + 1)

= (a) m × {m – (1 / m)} × {1 / (m + 1)}

[Using (am)n = amn]

Now,

Consider,

m × {m – (1 / m)} × {1 / (m + 1)}

= (m2 – 1) × {1 / (m + 1)}

= m2 × {1 / (m + 1)} – 1 × {1 / (m + 1)}

= {m2 / (m + 1)} – {1 / (m + 1)}

= {(m2 – 1)} / {(m + 1)}

= {(m – 1) (m + 1)} / (m + 1)

We get,

= (m – 1)

Therefore, (a) m × {m – (1 / m)} × {1 / (m + 1)} = am – 1

(h) xm + 2n. x3m – 8n ÷ x5m – 60

= xm + 2n + 3m – 8n – 5m – (- 60)

[Using am × an = am + n and am ÷ an = am – n]

= xm + 2n + 3m – 8n – 5m + 60

We get,

= x– m – 6n + 60

(i) (81)3 / 4 – (1 / 32)– 2 / 5 + 81 / 3. (1 / 2)-1. 20

This can be written as,

= (34)(3 / 4) – (1 / 25)(- 2 / 5) + (23)(1 / 3). (1 / 2)– 1 × 1

[Using a0 = 1]

= 34 × (3 / 4) – {1 / 25 × (- 2 / 5)} + 23 × (1 / 3). (2)1

= 33 – {1 / 2– 2} + 21. (2)1

= 33 – 22 + 2(1 + 1)

[Using am × an = am + n and a– n = (1 / an)]

= 33 – 22 + 22

We get,

= 33

= 27

(j) (27/343)2/3 ÷ {1/ (625/1296)1/4} x 536 / This can be written as,

= {33 / (73}2/3 ÷ [1/{54 / (24 x 34)}1/4] x (23 x 67) / = (33/ 73)2/3 ÷ [1/ {54 / (24 x 34)}1/4] x (23 x 67) / (33)1/3

= (33 x 2/3 / 73 x 2/3) ÷ [1/ {54 x ¼ / (24 x ¼ x 34 x ¼)] x (23 x 67) / (33 x 1/ 3)

Using (am)n = am n

= (32 / 72) ÷ [1 / {51 / (21 x 31)}] x (23 x 67) / 31

= (32 / 72) ÷ {(21 x 31) / 51} x (23 x 67) / 31

= (32 / 72) x {51 / (21 x 31)} x (23 x 67) / 31

On further calculation, we get,

= 32 – 1 – 1 x 23 – 1 x 51 x 7-2 x 67

Using a am ÷ an = am – n

= 30 x 22 x 51 x 7-2 x 67

= 1 x 4 x 51x 7-2 x 67

= 1340 / 49

= 27.34

8. Simplify the following:

(i) (5x × 7 – 5x) / (5x + 2 – 5x + 1)

(ii) (3x + 1 + 3x) / (3x + 3 – 3x + 1)

(iii) (2m × 3 – 2m) / (2m + 4 – 2m + 1)

(iv) (5n + 2 – 6.5n + 1) / (13. 5n – 2. 5n + 1)

Solution:

(i) (5x × 7 – 5x) / (5x + 2 – 5x + 1)

On taking common terms, we get,

= {5x (7 – 1)} / {5x + 1 (5 – 1)}

= (5x – x – 1 × 6) / 4 [Using am ÷ an = am – n]

= (5– 1 × 6) / 4

= {6 / (5 × 4)}

We get,

= (3 / 10)

(ii) (3x + 1 + 3x) / (3x + 3 – 3x + 1)

On taking common terms, we get,

= {3x (3 + 1)} / {3x (33 – 3)}

= {4 / (27 – 3)}

= (4 / 24)

We get,

= (1 / 6)

(iii) (2m × 3 – 2m) / (2m + 4 – 2m + 1)

On taking common terms, we get,

= {2m (3 – 1)} / {2m (24 – 2)}

= {2 / (16 – 2)}

= (2 / 14)

We get,

= (1 / 7)

(iv) (5n + 2 – 6.5n + 1) / (13. 5n – 2. 5n + 1)

On taking common terms, we get,

= {5n (52 – 6 × 5)} / {5n (13 – 2 × 5)}

= (25 – 30) / (13 – 10)

We get,

= (- 5 / 3)

9. Solve for x:

(a) 22x + 1 = 8

(b) 3 × 7x = 7 × 3x

(c) 2x + 3 + 2x + 1 = 320

(d) 9 × 3x = (27)2x – 5

(e) 22x + 3 – 9 × 2x + 1 = 0

(f) 1 = px

(g) p3 × p– 2 = px

(h) p– 5 = (1 / px + 1)

(i) 22x + 2x + 2 – 4 × 23 = 0

(j) 9 x 81x = 1/ 27x – 3

(k) 22x – 1 – 9 × 2x – 2 + 1 = 0

(l) : 5x = 25: 1

(m) = (0.6)2 – 3x

(n) = (27– 1) / (125– 1)

(o) 9x + 4 = 32 × (27)x + 1

Solution:

(a) 22x + 1 = 8

This can be written as,

22x + 1 = 23

2x + 1 = 3

2x = 3 – 1

2x = 2

We get,

x = 1

(b) 3 × 7x = 7 × 3x

(7x / 7) = (3x / 3)

7x – 1 = 3x – 1 [Using am ÷ an = am – n]

7x – 1= 3x – 1 × 1

7x – 1 = 3x – 1 × 70

[Using a0 = 1]

x – 1 = 0

We get,

x = 1

(c) 2x + 3 + 2x + 1 = 320

This can be written as,

2x + 3 + 2x + 1 = 26 × 5

2x. 23 + 2x.21 = 26 × 5

On taking common terms, we get,

2x (23 + 21) = 26 × 5

2x (8 + 2) = 26 × 5

2x (10) = 26 × 5

2x (10 / 5) = 26

2x. 2 = 26

(2x.2) / 26 = 1

2x + 1 – 6 = 1 × 20

2x – 5 = 1 × 20

x – 5 = 0

We get,

x = 5

(d) 9 × 3x = (27)2x – 5

This can be written as,

32 × 3x = (33)2x – 5

32 × 3x = 33 × (2x – 5)

On further calculation, we get,

32 + x = 36x – 15

1 = (36x – 15) / (32 + x)

1 = 36x – 15 – 2 – x

30 = 35x – 17

5x – 17 = 0

5x = 17

We get,

x = (17 / 5)

(e) 22x + 3 – 9 × 2x + 1 = 0

This can be written as,

22x.23 – 9 × 2x + 1 = 0

Put 2x = t

Then,

22x = t2

So,

22x.23 – 9 × 2x + 1 = 0 becomes,

8t2 – 9t + 1 = 0

8t2 – 8t – 1t + 1 = 0

On taking common terms, we get,

8t (t – 1) – 1 (t – 1) = 0

(t – 1) = 0 or (8t – 1) = 0

t = 1 or t = (1 / 8)

2x = 1 or 2x = (1 / 23)

2x = 20 or 2x = 2-3

Hence,

x = 0 or x = – 3

(f) 1 = px

p0 = px [Using a0 = 1]

Therefore,

x = 0

(g) p3 × p– 2 = px

p3 + (- 2) = px [Using am × an = am + n]

p3 – 2 = px

p1 = px

Hence,

x = 1

(h) p– 5 = (1 / px + 1)

p-5 × px + 1 = 1

p– 5 + x + 1 = 1 [Using am × an = am + n]

px – 4 = p0

x – 4 = 0

We get,

x = 4

(i) 22x + 2x + 2 – 4 × 23 = 0

This can be written as,

22x + 2x + 2 – 22 × 23 = 0

22x + 2x.22 – 22 + 3 = 0 [Using am × an = am + n]

22x + 2x.22 – 25 = 0

22x + 2x. 4 – 32 = 0

Put 2x = t

Then, 22x = t2

22x + 2x.4 – 32 = 0 becomes,

t2 + 4t – 32 = 0

t2 + 8t – 4t – 32 = 0

On taking common terms, we get,

t (t + 8) – 4 (t + 8) = 0

t + 8 = 0 or t – 4 = 0

t = – 8 or t = 4

2x = – 8 or 2x = 4

2x = – 23 or 2x = 22

Now,

Consider second equation,

2x = 22

We get,

x = 2

(j) 9 x 81x = 1/ 27x – 3

This can be written as,

32 x 34x = 1/ 33 (x – 3)

32 x 34x = 1/ 33x – 9

Using (am)n = amn

32 x 34x x 33x – 9 = 1

32 + 4x + 3x – 9 = 1 x 30

On further calculation, we get,

2 + 4x + 3x – 9 = 0

7x – 7 = 0

7x = 7

x = 1

(k) 22x – 1 – 9 × 2x – 2 + 1 = 0

This can be written as,

22x. 2-1 – 9 × 2x. 2– 2 + 1 = 0

Let 2x = t,

So, 22x = t2

Then,

22x.2– 1 – 9 × 2x.2– 2 + 1 = 0 becomes,

(t2 / 2) – 9 × (t / 22) + 1 = 0

(t2 / 2) – (9t / 4) + 1 = 0

Taking LCM, we get,

2t2 – 9t + 4 = 0

2t2 – 8t – t + 4 = 0

2t (t – 4) – 1 (t – 4) = 0

t – 4 = 0 or 2t – 1 = 0

t = 4 or t = (1 / 2)

Hence,

2x = 4 or 2x = (1 / 2)

2x = 22 or 2x = 2– 1

Therefore,

x = 2 or x = – 1

(l) : 5x = 25: 1

This can be written as,

( / 5x) = (25 / 1)

( / 5x) = (52 / 1)

= 52 × 5x = 52 + x x2 = 2 + x

x2 – x – 2 = 0

(x – 2) (x + 1) = 0

x – 2 = 0 or x + 1 = 0

Therefore,

x = 2 or x = – 1

(m) = (0.6)2 – 3x

This can be written as,

{1 + (2 / 3)}1 / 2 = (6 / 10)2 – 3x

On taking LCM, we get,

(5 / 3)1 / 2 = (3 / 5)2 – 3x

(3 / 5)– 1 / 2 = (3 / 5)2 – 3x

(- 1 / 2) = 2 – 3x

-1 = 2 (2 – 3x)

– 1 = 4 – 6x

– 1 – 4 = – 6x

– 5 = – 6x

Hence,

x = (5 / 6)

(n) = (27– 1) / (125– 1)

This can be written as,

(3 / 5)(x + 3) × (1 / 2) = {(33)-1 / (53)-1}

(3 / 5)(x + 3) / 2 = (3 / 5)– 3

(x + 3) / 2 = – 3

x + 3 = – 3 × 2

x + 3 = – 6

x = – 6 – 3

We get,

x = – 9

(o) 9x + 4 = 32 × (27)x + 1

This can be written as,

9x + 4 = 32 × (33)x + 1

32 (x + 4) = 32 × 33x + 3

32x + 8 = 32 + 3x + 3

Hence,

2x + 8 = 2 + 3x + 3

2x + 8 = 3x + 5

3x – 2x = 8 – 5

We get,

x = 3

10. Find the value of k in each of the following:

(i) = 2k

(ii) = xk

(iii) × = 3k

(iv) (1 / 3)– 4 ÷ 9 (- 1 / 3) = 3k

Solution:

(i) = 2k

This can be written as,

8(1 / 3) × (- 1 / 2) = 2k

(23)(1 / 3) × (- 1 / 2) = 2k

2(- 1 / 2) = 2k

Therefore,

k = (- 1 / 2)

(ii) = xk

This can be written as,

{(x2)1 / 3}1 / 4 = xk

On further calculation, we get,

(x2)1 / 12 = xk

x(2 / 12) = xk

x(1 / 6) = xk

Hence,

k = (1 / 6)

(iii) × = 3k

This can be written as,

{(32)(1 / 2)}– 7 × {(3)(1 / 2)}– 5 = 3k

3– 7 × 3(- 5 / 2) = 3k

3-7 – 5 / 2 = 3k

3(- 14 – 5) / 2 = 3k

3(- 19 / 2) = 3k

Therefore,

k = (- 19 / 2)

(iv) (1 / 3)– 4 ÷ 9 (- 1 / 3) = 3k

This can be written as,

(3– 1)– 4 ÷ (32)– 1 / 3 = 3k

34 ÷ 3(- 2 / 3) = 3k

34 – (- 2 / 3) = 3k

34 + 2 / 3 = 3k

3(14 / 3) = 3k

We get,

k = (14 / 3)

11. If a = 2(1 / 3) – 2(- 1 / 3), prove that 2a3 + 6a = 3

Solution:

Given

a = 2(1 / 3) – 2(- 1 / 3)

This can be written as,

a = 2(1 / 3) – {1 / 2(1 / 3)}

On taking cube on both sides, we get,

a3 = [2(1 / 3) – {1 / 2(1 / 3)}]3

On further calculation, we get,

a3 = 2 – (1 / 2) – 3 [2(1 / 3) – {1 / 2(1 / 3)}]

a3 = {(4 – 1) / 2} – 3a

a3 = (3 / 2) – 3a

We get,

2a3 + 6a = 3

12. If x = 3(2 / 3) + 3(1 / 3), prove that x3 – 9x – 12 = 0

Solution:

Given

x = 32 / 3 + 31 / 3

x3 = 32 + 3 + 3 × 32 / 3 × 31 / 3 (32 / 3 + 31 / 3)

x3 = 9 + 3 + 3 × 32 / 3 + 1 / 3 (x)

x3 = 12 + 9x

We get,

x3 – 9x – 12 = 0

13. If and abc = 1, prove that x + y + z = 0

Solution:

Let = k

So,

a1 / x = k, b1 / y = k, c1 / z = k

We get,

a = kx, b = ky , c = kz

Also given that,

abc = 1

kx × ky × kz = 1

kx + y + z = k0

We get,

x + y + z = 0

14. If ax = by = cz and b2 = ac, prove that y = 2xz / (z + x).

Solution:

Let ax = by = cz = k

So,

a = k1 / x, b = k1 / y, c = k1 / z

Also given that,

b2 = ac

k2 / y = k1 / x × k1 / z

k2 / y = k1 / x + 1 / z

(2 / y) = (1 / x) + (1 / z)

(2 / y) = (z + x) / zx

We get,

y = {2zx / (z + x)}

15. Show that: {1 / (1 + ap – q)} + {1 / (1 + aq – p)} = 1

Solution:

Consider LHS of the equation, i.e,

{1 / (1 + ap – q)} + {1 / (1 + aq – p)}

On taking LCM, we get,

= {(1 + aq – p) + (1 + ap – q)} / (1 + ap – q) (1 + aq – p)

= (2 + a– (p – q) + ap – q) / (1 + ap – q) (1 + a– (p – q))

= 2 + a– (p – q) + ap – q / (1 + a– (p – q) + ap – q + ap – q. a– (p – q))

= 2 + a– (p – q) + ap – q / (1 + a– (p – q) + ap – q + ap – q – p + q

= 2 + a– (p – q) + ap – q / (1 + a– (p – q) + ap – q + a0)

= 2 + a– (p – q) + ap – q / (1 + a– (p – q) + ap – q + 1)

= 2 + a– (p – q) + ap – q / (2 + a– (p – q) + ap – q)

We get,

= 1

= RHS

Therefore,

LHS = RHS

Hence, proved