Frank Solutions for Class 9 Maths Chapter 9 Indices

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frank solutions class 9 maths chapter 9 01
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frank solutions class 9 maths chapter 9 11
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Access Frank Solutions for Class 9 Maths Chapter 9 Indices

1. Evaluate the following:

(i) 60

(ii) (1 / 2)-3

(iii) FRANK Solutions Class 9 Maths Chapter 9 - 1

(iv) FRANK Solutions Class 9 Maths Chapter 9 - 2

(v) (0.008)(2 / 3)

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

(vii) FRANK Solutions Class 9 Maths Chapter 9 - 3

(viii) FRANK Solutions Class 9 Maths Chapter 9 - 4

Solution:

(i) 60 = 1

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

We get,

(1 / 2)-3 = 8

(iii)
FRANK Solutions Class 9 Maths Chapter 9 - 5= 26

We get,

= 64

 

(iv)
FRANK Solutions Class 9 Maths Chapter 9 - 7= 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)
FRANK Solutions Class 9 Maths Chapter 9 - 9=
FRANK Solutions Class 9 Maths Chapter 9 - 10

=
FRANK Solutions Class 9 Maths Chapter 9 - 11

= 56 × 1 / 6

We get,

= 5

(viii)
FRANK Solutions Class 9 Maths Chapter 9 - 13= (64 / 27)(2 / 3)

= (4 / 3)3 × 2 / 3

= (4 / 3)2

We get,

FRANK Solutions Class 9 Maths Chapter 9 - 14= (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) FRANK Solutions Class 9 Maths Chapter 9 - 15

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)
FRANK Solutions Class 9 Maths Chapter 9 - 16

= (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) FRANK Solutions Class 9 Maths Chapter 9 - 17 ÷ FRANK Solutions Class 9 Maths Chapter 9 - 18

(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 / FRANK Solutions Class 9 Maths Chapter 9 - 19

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)
FRANK Solutions Class 9 Maths Chapter 9 - 20÷
FRANK Solutions Class 9 Maths Chapter 9 - 21

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,

=
FRANK Solutions Class 9 Maths Chapter 9 - 22

(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 /
FRANK Solutions Class 9 Maths Chapter 9 - 23

This can be written as,

= {33 / (73}2/3 ÷ [1/{54 / (24 x 34)}1/4] x (23 x 67) /
FRANK Solutions Class 9 Maths Chapter 9 - 24

= (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) FRANK Solutions Class 9 Maths Chapter 9 - 25: 5x = 25: 1

(m) FRANK Solutions Class 9 Maths Chapter 9 - 26 = (0.6)2 – 3x

(n) FRANK Solutions Class 9 Maths Chapter 9 - 27 = (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)
FRANK Solutions Class 9 Maths Chapter 9 - 28: 5x = 25: 1

This can be written as,

(
FRANK Solutions Class 9 Maths Chapter 9 - 29/ 5x) = (25 / 1)

(
FRANK Solutions Class 9 Maths Chapter 9 - 30/ 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)
FRANK Solutions Class 9 Maths Chapter 9 - 33= (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)
FRANK Solutions Class 9 Maths Chapter 9 - 34 = (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) FRANK Solutions Class 9 Maths Chapter 9 - 35= 2k

(ii) FRANK Solutions Class 9 Maths Chapter 9 - 36 = xk

(iii) FRANK Solutions Class 9 Maths Chapter 9 - 37× FRANK Solutions Class 9 Maths Chapter 9 - 38= 3k

(iv) (1 / 3)– 4 ÷ 9 (- 1 / 3) = 3k

Solution:

(i)
FRANK Solutions Class 9 Maths Chapter 9 - 39= 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)
FRANK Solutions Class 9 Maths Chapter 9 - 40= 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)
FRANK Solutions Class 9 Maths Chapter 9 - 41×
FRANK Solutions Class 9 Maths Chapter 9 - 42= 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 FRANK Solutions Class 9 Maths Chapter 9 - 43 and abc = 1, prove that x + y + z = 0

Solution:

Let
FRANK Solutions Class 9 Maths Chapter 9 - 44= 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

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