# OP Malhotra Class 9 ICSE Maths Solutions Chapter 6

## OP Malhotra Class 9 ICSE Maths Solutions Chapter 6 Indices/Exponents

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 6, Indices/Exponents. Here students can easily find step by step solutions of all the problems for Indices/Exponents, Exercise 6a, 6b, 6c and 6d Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 6 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Indices/Exponents Exercise 6(a) Solution :

Question no – (1)

Solution :

(i) (x. x. x. x. x) (x . x)

= x5.x2

= x5+2

= x7

Thus, the product in exponential form x7

(ii) -1 (n . n . n) (n . n)

= – n3.n2

= – n5

Thus, the product in exponential form -n5

(iii) y8 × y5 × y2

= y8 + 5 + 2

= y15

Hence, the product in exponential form y15

(iv) -x (- x5)

= x6

Thus, the product in exponential form x6

(v) (3a7b8c9) × (5a27b16c8)

= 15a(27 + 7) b(8+16). c (9+8)

= 15a34 b24 c17

So, the product

in exponential form is 15a34 b24 c17

(vi) (x + 2)2 . (x + 2)4

= (x + 2)6

So, the product in exponential form is (x + 2)6

(vii) (2m – 3n)3a-2b × (2m – 3n)6a + 10b

= (2m – 3n)3a – 2b × (2m – 3n)6a+10b

= (2m – 3n)3a-2b+6a+10b

= (2m – 3n)9a+8b

Thus, the product in exponential form is (2m – 3n)9a+8b

(viii) x2a+b-c. x2c+a-b. x2b+c-a

= x2a+b+c+2c+a-b+2b+c-a

= x2a+2b+2c

Therefore, the product in exponential form is x2a+2b+2c

Question no – (2)

Solution :

(i) (xy)3

= x3. y3…(Simpler form)

(ii) (- x)5

= (-x). (-x), (-x), (-x), (-x), (-x)…(Simpler form)

(iii) (- 2xy)4

= 16x4y4…(Simpler form)

(iv) (p/q)8

= p8/q8…(Simpler form)

(v) (x2)5

= x10…(Simpler form)

(vi) (73)8

= 724…(Simpler form)

(vii) (6a2)3

= 216a5…(Simpler form)

(viii) (- x2y3)3

= – x6y9…(Simpler form)

(ix) (p2)5 × (p3)2

= p10. p6

= p16

(x) 3 (x4y3)10 × 5(x2y2)3

= 15x46. y36…(Simpler form)

(xi) (c3/d2)7

= c21/d14…(Simpler form)

(xii) (3p2/4q2)a

= 3ap2a/4aq2a…(Simpler form)

(xiii) (a2b2/x2y3)m

= a2mb2m/x2m.y3m…(Simpler form)

Question no – (3)

Solution :

(i) x6 ÷ x2

= x6-2

= x4

Thus, here the quotient is x4

(ii) x2a ÷ xa

= xa

The quotient is = xa

(iii) p5q3/p3q2

= p5-3. q3-2

= p2.q1

Thus, here the quotient will be p2.q1

(iv) -35x10y5/-7 x3y2

Now, -35x10y5/-7 x3y2

= 5x7y3

Thus, here the quotient is 5x7y3

(v) (- 8 x27y21) ÷ (- 16 x6y17)]

Now, (- 8 x27y21) ÷ (- 16 x6y17)]

= 1.x21y4/2

Thus, here the quotient is 1.x21y4/2

(vi) 4 pq2 (- 5pq3)/10p2q2

= 2 (- 5p2q3)/5p2

= – 2q3

Thus, here the quotient is -2q3

(vii) (-4ab2)2/16ab

= + 16a2b4/16ab

= + ab3

Thus, here the quotient is + ab3

(viii) xa-b yc-d/x2b-a yc

= xa-b-2b+a yc-d –c

= x2a-3b y-d

Thus, here the quotient is x2a-3b y-d

(ix) (m3n-9)6/m2n-4

= m18-54/m2n-4

= m18n-54 – 2n + 4

= m16x-50

Thus, here the quotient is m16x-50

(x) [(x2a-4)2/xa+5]3

= [x4a-8/xa+5]3

= x12a – 24/x3a+15

= x12a-24 – 3a – 15

= x9a – 39

Therefore, here the quotient is x9a – 39

Indices/Exponents Exercise 6(b) Solution :

Question no – (1)

Solution :

(i) 50

= 1

(ii) 2-3

= 1/32

= 1/8

(iii) 10-4

= 1/104

= 1/10000

(iv) (-2)-2

= 1/(-2)2

= 1/4

(v) (2/3)0

= 1

(vi) (3/4)-3

= (4/3)3

= 64/27

(vii) (- 1/2)-2

= (- 2)2

= 4

Question no – (2)

Solution :

(i) 41/2

= (22)1/2

= 22×1/2

= 2

(ii) 81/3

= (2)3×1/3

= 2

(iii) 161/4

= (4)2×1/4

= 41/2

= (2)2×1/2

= 2

(iv) 271/3

= (-3)3×1/3

= (-3)1

= – 3

(v) 323/5

= (2)5×3/5

= (2)3

= 8

(vi) (125)-2/3

= (5)3×(/3)

= (-5)2

= 1/52

= 1/25

(vii) (8/125)1/3

= (2/5)3×1/3

= 2/5

(viii) (1/216)-2/3

= (1/6)3×(-=2/3)

= (1/6)-2

= 36

(ix) – (-27) -4/3

= – 1/(-3)4

= – 1/(-3)4

= – 1/81

(x) (27/8)-2/3

= (3/2)3×(-2/3)

= (3/2)-2

= (2/3)2

= 4/9

Indices/Exponents Exercise 6(c) Solution :

Question no – (1)

Solution :

(i) Given, (12 19/27)1/3

= (12 × 27 + 19/27)1/3

= (324 + 19/27)1/3

= (343/27)1/3

= (7/3)3×1/3

= 7/3

(ii) 3√(64)-4 (125)-2

Now, 3√(64)-4 (125)-2

= 3√(64) -4 (125) -2

= (43-4/3 . (53– 2

= (4) – 4 . (5) – 2

= 1/4 × 4 × 4 × 4 × 5 × 5

= 1/6400

(iii) (9-3 × 16 2/31/6

= [(3) -3 × (423/21/6

= 3 – 6 × 1/6 × (4) 3 × 1/6

= 3 -1 × 4 1/2

= 1/3 × 2

= 2/3

(iv) 3√(16)-3/4 × (125)-2

= 3√(2) 4 × -3/4 (53-2

= [1/23 . 56]3

= 1/2 × 52

= 1/50

(v) (32) -2/5 ÷ (216) -2/3

= (2)5 × -2/5 ÷ (6) 3 × -2/3

= 2 -2 ÷ 6 -2

= (2/6) – 2

= (6/2) 2

= 36/4

= 9

Question no – (2)

Solution :

(i) Given, (64/125)-2/3 ÷ 1/(256/625)1/4 + (√25/3√64)0

= (125/64) 2/3 × (4/5) 4 × 1/4 + 1

= (5/4) 3 × 2/3 × (4/5) + 1

= (5/4) 2 × 4/5 + 1

= (5/4) 2 × (5/4) – 1 + 1

= (5/4) 2 – 1 + 1

= 5/4 + 1

= 5 + 4/1

= 9/4

= 2 1/4

(ii) 93/2 – 3 × (5)0 – (1/81)-1/2

= (3)2×3/2 – 3 × 1 – (81)1/2

= 33 – 3 – (9)2×1/2

= 27 – 3 – 9

= 15

(iii) (1/4)-2 – 3(8)2/3 × 40 + (9/16)-1/2

∴ (1/4)-2 – 3(8)2/3 × 40 + (9/16)-1/2

= (4)2 – 3 × (2)3×2/3 × 1 + (9/16)-1/2

= 16 – 12 + (3/4)2×1/2

= 16 – 12 + 4/3

= 4 + 4/3

= 5 1/3

(iv) 163/4 + 2(1/2)-1 (3)0

∴ 163/4 + 2(1/2)-1 (3)0

= (2)4×3/4 + 2 × 2 × 1

= 23 + 4

= 8 + 4

= 12

(v) (81)3/4 – (1/32)-2/5 + (8)1/3 (1/2)-2 (2)0

Now, (81)3/4 – (1/32)-2/5 + (8)1/3 (1/2)-2 (2)0

= (3)4×3/4 – (32)5/2 + (2)3×1/3 × 22 × 1

= 33 – (1/25)-2/5 + (2)3×1/3 × 22 × 1

= 33 – (1/2)5×(-2/5) + 2 × 4

= 27 – 4 + 8

= 31

Question no – (3)

Solution :

(i) As per the question we know,

x = 9,

y = 2,

z = 8

Now, 91/2 × (2)-1 × (8)2/3

= (3)2×1/2 × 1/2 × (2)3×2/3

= 3 × 1/2 × 22

= 6

(ii) (27/8)2/3 – (1/4)-2 + (5)0

∴ (27/8)2/3 – (1/4)-2 + (5)0

= (2/3)8×2/3 – (1/2)2×(-2)

= 4/9 – 24 + 1

= 9/4 – 16 + 1

= – 12 3/4

(iii) √1/4 + (0.01)-1/2 – (27)2/3

Now, (1/22)1/2 + (0.1)2×(-1/2) – (3)3×(-2/3)

= 1/2 + 1/0.1 – 9

= 1/2 + 10 – 9

= 1 1/2

Question no – (4)

Solution :

(i) In the question, (64) 5n/6 . (27) –n/6/(12) –n/2 . 26n

Now, (64) 5n/6 . (27) –n/6/(12) –n/2 . 26n

= (2)6×5n/6 (27)-n/6/(22×3)-n/2 × 26n

= 25n 3-n/2/2-n 3-n/2 × 26n

= 25n+n+6n. 3-n/2+n/2

= 212n .3

= 112n…(Simplified)

(ii) 6 . (8) n + 1 + 16 . 2 3n – 2/10 . 2 3n + 1 – 7 . (8)n

∴ 6 . (8) n + 1 + 16 . 2 3n – 2/10 . 2 3n + 1 – 7 . (8)n

= 6.(2)3n+3 + 16.23n-2/10.23n+1 – 7(2)3n

= 6.23n.23 + 16.23n.2-2/10.23n .2-(7.2)3n

= 23n (6 × 23 + 16 × 2-2)/23n (10 × 2 – 7)…(Simplified)

(iii) 42n . 2n + 1/2n – 3. 42n + 1

∴ 42n . 2n + 1/2n – 3 . 42n + 1

= (22)2n. 2n+1/2n-3. (22)2n+1

= 24n. 2n+1/2x-3 . 24n+2

= 24n+n+1/2n-3+4n+2

= 25n+1/5n-1

= 25n+ -5n+1

= 22

= 4…(Simplified)

(iv) (5)2n + 3 – (25)n + 2/[(125)n + 1]2/3

∴ (5)2n + 3 – (25)n + 2/[(125)n + 1]2/3

= 52n .53 – (52)n+2/[(53)n+1]2/3

= 52n , 53 – 52n.54/52n .52

= 52n (53 – 54)/52n. 25

= 125 – 625/25

= – 500/25

= – 20…(Simplified)

Question no (5)

Solution :

According to the question, 49n+1 . 7n-(343)n/73m 24 = 3/343

∴ (72)n+1. 7n – (73)n/73n .2= 3/343

= 72n+2. 7n-73n/73m .24 – 3/343 = 0

= 73n (72 – 1)/73m . 16 = 3/343

= 73n, (72 – 1)/73m . 16 = 3/343

= 73n . 46/73m. 16 = 3/343

= 73n/73m = 3 × 16/343 × 48 = 1/343

= (1/7)3

= 7n/7m = 1/7

= 7,7n = 1/7

= 7n+1 = 7m

= m = n + 1 …(Proved)

Question no (6)

Solution :

(i) As per the question we get, 2205 = 3a × 5b × 7c,

∴ 2205 = 3a × 5b × 7c,

= 51 × 32 × 72 = 2025

Comparing with 3a × 5b × 7c

Here, a = 2, b = 1, c = 2

Therefore, the values of a = 2, b = 1 and c = 2

(ii) 3a × 5-b × 7-c

∴ 3a × 5-b × 7-c

= 3a/5b × 7c

= 32/51 × 72

= 9/49 × 5

= 9/245

Question no – (7)

Solution :

Need to prove, x + y + 6 = 0

= (b3c-2/b-4 c3)-3 ÷ (b-1c/b2c-2)= bx cy

= b-9. c6/b12. c-9 ÷ b-5. c-5/b10 c-10 = bx cy

= b-9 – 12 + 10 + 5 C6+9-10-5 = bx cy

= b -21+15 C15-10-5 = bx cy

= b-6 c0 = bx cy

= So, x = – 6, y = 0

L.H.S,

Now, x + y + 6

= – 6 + 0 + 6

= 0

L.H.S = R.H.S ….(Proved)

Question no – (8)

Solution :

(i) (x + y)-1 (x-1 + y-1) = 1/xy

L.H.S (x + y)-1 (x-1 + y-1) = 1/xy

= 1/x + y × (x-1 + y-1)

= 1/xy × (1/x + 1/y)

= (1/(x + y) × (y + x)/xy

= 1/xy

L.H.S = R.H.S ….(Proved)

(ii) (x-1 + y-1)-1 = xy/x + y

L.H.S

= (x-1 + y-1)-1

= (1/x + 1/y)-1

= xy/x + y

Thus, L.H.S = R.H.S …(Proved)

(iii) a + b + c/a-1b-1 + b-1 c-1 + c-1 a-1 = abc

L.H.S, a + b + c/a/1 × 14/b + 1/b × 1/c + 1/c × 1/a

= a + b + c/c + a + b/abc

= (a + b + c) × abc/a+b+c

= abc

Hence, L.H.S = R.H.S …(Proved)

(iv) 1/1 + xb-a + xc-a + 1/1 + xa-b + xc-b + 1/1 + xb-c + xa-c = 1

∴ 1/a-a + xb-b + xc-a + xb-b + xa-b + xc-b + 1/xc-c + xb-c + xa-x

= 1/x-a (xa + xb + xc) + 1/x-b (xb + xa + xc) + 1/x-c (xc + xa + xb)

= 1/xa + xb + xc [1/x – a + 1/x – b + 1/x – c]

= 1/2a + xb + xc [xa + xb + xc/1]

= 1

L.H.S = R.H.S…(Proved)

(v) ab√xa/xb. bc√xb/xc. ca√xc/xa = 1

Now, ab√xa/xb. bc√xb/xc. ca√xc/xa = 1

= (x a-b)/ab . (xb-c)1/bc. (xc-a)/ca

= xa-b/ab. x b-c/bc. x c-a/ca/abc

= x0/abc

= x0

= 1

Therefore, L.H.S = R.H.S…(Proved)

Question no – (9)

Solution :

(i) (p1/3 –p-1/3) (p2/3 + 1 + p-2/3)

Let, x = p1/3, p/1/3 = y

∴ (x – y) (x2 + xy + y2)

= x3 – y3

= (p1/3)3 – (p-1/3)3

= p1 – p-1

= p-1/p…(Simplified)

(ii) (√11 + √3)1/3 (√11 – √3)1/2

Here, Let, √11 = x, √3 = y

= (x + y)1/3 (x – y)1/3

= (x2 – y2)1/3

= [(√11)2 – (√3)2]1/3

= (1 – 3)1/3

= (8)1/3

= (2)3×1/3

= 2 …(Simplified)

Question no – (11)

Solution :

Here, L.H.S,

= xq-r × yr-p. zp-q

= (ab)(p-1)(q-r) × (ab)(q-1)(r-p) × (ab)(p-q). (r-1)

= (ab)(pq – pr – q + r) (ab) (qr – pq – r + p) (ab(pr – p qr + q)

= (ab) (pq – pr – q + r + qr – pq – r + p + pr – p qr + q)

= (ab)c

= 1

Therefore, L.H.S = R.H.S …(Proved)

Indices/Exponents Exercise 6(D) Solution :

Question no – (1)

Solution :

Given equation, 2x+1 = 4x-3

Now, 2x+1 = 4x-3

= 2x+1 = (2)2(x – 3)

= (2)2n – 6

= x + 1 = 2n – 6

= x – 2x = – 6 – 1

= – x = – 7

= x = 7

Question no – (2)

Solution :

In the question, x-3/4 = 1/8

Now, x-3/4 = 1/8

= x-3/4 = 1/23

= 2-3

= (x 1/4) -3 = (2)-3

= x = 24

= 16

Question no – (3)

Solution :

In the given question, (x – 1)2/3 = 25

(x – 1)2/3 = 25

= (x – 1)2/3 = 52

= (x – 1)2/3 = (53)2/3

= x – 1 = 53

= x – 1 = 185

= x = 125 + 1

= 126

Question no – (4)

Solution :

Given in the question, 25x+3 = 8x+3

Now, 25x+3 = 8x+3

= 25x+3 = 23x+9

= 5x + 3 = 3x + 9

= 5x – 3x = 9 – 3

= x = 6/2

= 3

Question no – (5)

Solution :

From the question we get, (√5/7)x-1 = (125/343)-1

Now, (√5/7)x-1 = (125/343)-1

= (5/7)x-1/2 = (53/73)-1

= (5/7)x-1/2 = (5/7)-3

= x – 1/2 = – 3

= x – 1 = – 6

= x = – 6 – 1

= – 5

Question no – (6)

Solution :

Given, 113-4x = (√1/121)-2

Now, 113-4x = (√1/121)-2

= 113-4x (√1/(121)-2

= (11)-2x – 2/2

= 3x – 4x = 2

= – 4x = – 1

= x = 1/4

Question no – (7)

Solution :

As per the question, (3√4)2x+1/2 = 1/32

(3√4)2x+1/2 = 1/32

= (3√(2)2x+1/2 = 1/25

= (22/3)2x + 1/2 = 25

= 2 2/3 (2x + 1/2) = 2-5

2/3 (2x + 1/2) = – 5

= 4/3 x + 1/3 = – 5

= 4x/3 = – 5 – 1/*2

= 4x/3 = – 15- 1/3

= 4x = – 16

= 4x = – 16/4

= x = – 4

Question no – (8)

Solution :

In the given question, √p/q (q/p)1-2x

∴ (p/1)1/2 = (p/q)– (1 – 2x)

= 2x – 1 = 1/2

= 2x = – 1 = 1/2

= 2x = 1/2 + 1

= 3/2 × 1/2

= 3/4

Hence the value of x will be 3/4

Question no – (9)

Solution :

Given, 23 (50 + 32x) = 8 8/27

Now, 23 (50 + 32x) = 8 8/27

= 8 (1 + 32x) = 224/27

= (1 + 22x) = 224/27 × 1/8 = 28/27

= 32x = 28/27 – 1

= 28 – 27/27

= 1/27

= 3-3

= 2x = -3

= x =  – 3/2

Therefore, the value of x will be -3/2

Question no – (11)

Solution :

Given in the question, 32x+4 + 1 = 2.3x + 2

= 32x. 34+1 = 2.3x .32 = 18.3x

= 81.32x + 1 = 18.3x

= 81.32x – 18.3x + 1= 0

= 3x = a

= 32x = a2

81a – 18a + 1 = 0

= (9)2 – 2,9 – 1 + 12 = 0

= (9a – 1)2 = 0

= 9a = 1

= a = 1/9

= 3x = 1/9

= 3x = 1/32

= 3-2

= x = – 2

Question no – (12)

Solution :

According to the question,

= 52x+1 = 6.5x –1

= 52x-1 = 6.5x + 1 = 0

= 5.52x – 6,5x + 1 = 0

Let, 5x = a, 52x = a2

5a2 – 6a + 1 = 0

= 5a2 – 5a – a + 1 =

= 5a (a – 1) (a – 1) = 0

5a – 1 = 0 and a – 1 = 0

= a = 1

= a = 1, 5x = 1 = 50

= x = 0

and, a = 1/5, 5x

= 1/5 = 5-1

= x = – 1

Question no – (13)

Solution :

Given,2x2 – 2x+3 = – 24

= 22x – 3x 23 + 24 + 24 = 0

= 22x + 16 – 2.2x = 0

Let, 2x = a

= 22x = a2

= a2 – 8a + 16 = 0

= a2 – 2a.4 + 42 = 0

= (a – 4)= 0

= a – 4 = 0

= a = 4

2= 4

= 2x = 22

= x = 2

Question no – (14)

Solution :

In the question we get, 9x = 3y-2, 81y = 32 × (27)x

= {(3)2}x = 3y -2

= 3x2 = 3y-2

= 2x = y – 2

= y = 2x + 2

-81y = 32 × (27)x

= (34)y = 32 × 33x

= 4y = 2 + 3x

Now, (2x + 2) = 3x + 2

= 8x + 8 = 3x + 2

= 8x + 8 = 3x + 2

= 8x – 3x = + 2 – 8

= 5x = – 6

= x = – 6/5

y = 2x + 2

= 2 × (-6/5) + 2

= – 12/5 + 2

= – 12 + 10/5

= – 2/5

Question no – (16)

Solution :

Given in the question, 2x = 16 ×2y, (27)x = 9 × 32y

Now, 2x = 16 × 2y

= 24 × 2y

= 24 × 2y

= 2x = 24 + y

= x = 4 + y – (i)

and, (27)x = 9 × 32y

= (33)x = 32 + 32y

= 33x = 3

3 (4 + y) = 2 + 2y

= 12 + 3y = 2 + 2y

= 3y – 2y = 2 – 12

= y = 10 = x = 4 + y

= 4 – 10

= – 6

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Updated: June 19, 2023 — 3:45 pm