# Selina Concise Solutions for Chapter 5 Exponents (including Laws of Exponents) Class 7 ICSE Mathematics

### Exercise 5 (A)

1. Find the value of:

(i) 6

(ii) 7

(iii) 4

(iv) 5

(v) 8

(vi) 7

(i) 6  = 6 × 6 = 36

(ii) 7  = 7 × 7 × 7 = 343

(iii) 4  = 4 × 4 × 4 × 4 = 256

(iv) 5 = 5 × 5 × 5 × 5 × 5 = 3125

(v) 8  = 8 × 8 × 8 = 512

(vi) 7 = 7 × 7 × 7 × 7 × 7 =16807

2.Evaluate:

(i) 2 ×  4

(ii) 2 ×  5

(iii) 3 ×  5

(iv) 2 ×  3

(v) 3 ×  5

(vi) 5×  2

(vii) 3×  4

(viii) (4 × 3)

(ix) (5 × 4)

(i) 2 ×  4

= 2 × 2 × 2 × 4 × 4

= 8 × 16

= 128

(ii) 2 ×  52

= 2 × 2 × 2 × 5 × 5

= 8 × 25

= 200

(iii) 3 ×  5

= 3 × 3 × 3 × 5 × 5

= 27 × 25

= 675

(iv) 2 ×  3

= 2 × 2 × 3 × 3 × 3

= 4 × 27

= 108

(v) 3 ×  5

=3 × 3 × 5 × 5 × 5

= 9 × 125

= 1125

(vi) 5×  2

= 5 × 5 × 5 × 2 × 2 × 2 × 2

= 125 × 16

= 2000

(vii) 3×  4

=3 × 3 × 4 × 4

= 9 × 16

=144

(viii) (4 × 3)

=4 × 4 × 4 × 3 × 3 × 3

= 64 × 27

= 1728

(ix) (5 × 4)

=5 × 5 × 4 × 4

= 25 × 16

= 400

3. Evaluate :

(i) (3/4)

(ii) (-5/6)

(iii) (-3/-5)

4. Evaluate :

(i) (2/3) × (3/4)

(ii) (-3/4) × (2/3)

(iii) (3/5) × (-2/3)

5. Which is greater :

(i) 2 or 3

(ii) 2 or 5

(iii) 4 or 3

(iv) 5 or 4

(i) 2 or 3

Since, 2  = 2 × 2 × 2 = 8

and, 32 = 3 × 3 = 9

∵ 9 is greater than 8

⇒ 32  > 2

(ii) 2 or 5

Since, 25 = 2 × 2 × 2 × 2 × 2 = 32

and, 52 = 5 × 5 = 25

∵ 32 is greater than 25

⇒ 25  > 52

(iii) 4 or 3

Since, 4  = 4 × 4 × 4 = 64

and, 3 = 3 × 3 × 3 × 3 = 81

∵ 81 is greater than 64

⇒ 3  > 4

(iv) 5 or 4

Since, 5  = 5 × 5 × 5 × 5 = 625

and, 4= 4 × 4 × 4 × 4 × 4= 1024

∵ 1024 is greater than 625

⇒ 4 >  5

6. Express each of the following in exponential form :

(i) 512

(ii) 1250

(iii) 1458

(iv) 3600

(v) 1350

(vi) 1176

7. If a = 2 and b = 3, find the value of:
(i) (a + b)2
(ii) (b – a)3
(iii) (a × b)a
(iv) (a × b)b
(i) (a + b)2
= (2 + 3)2
= (5)2
= 5 × 5 = 25

(ii) (b – a)2
= (3 – 2)2
= (1)3
= 1 × 1 × 1
= 1

(iii) (a × b)a
= (2 × 3)2
= (6)2
= 6 × 6
= 36

(iv) (a × b)b
= (2 × 3)3
= (6)3
= 6 × 6 × 6
= 216

8. Express:
(i) 1024 as a power of 2.
(ii) 343 as a power of 7.
(iii) 729 as a power of 3.
(i) 1024 as a power of 2.

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 210

9. If 27 × 32 = 3x × 2y ; find the values of x and y.
27 × 32 = 3x × 2y
⇒ 27 = 3x
⇒ 27 = 3×3×3
⇒ 3= 3x
∴ x = 3
Also, 32 = 2y

32 = 2 ×2×2×2×2
⇒ 25 = 2y
∴ y = 5

10. If  64 × 625 = 2a × 5b ; find :
(i) the values of a and b.
(ii) 2b × 5a
(i) the values of a and b.
64 × 625 = 2a × 5b
⇒ 64 = 2a
⇒ 64 = 2×2×2×2×2×2
⇒ 64 = 26
∴ a = 6
Also, 625 = 5b

625 = 5×5×5×5
⇒ 625 = 54
∴ b = 4

(ii) 2b × 5a
= 24 × 56
= 2×2×2×2×5×5×5×5×5×5
= 16 × 15625
= 250000

### Exercise 5 (B)

1. Fill in the blanks:
In 52 = 25, base = ......... and index = ..........
If index = 3x and base = 2y, the number = .........
(i) In 52  = 25, base = 5 and index = 2
(ii) If index = 3x and base = 2y, the number = 2y3x

2. Evaluate :
(i) 28 ÷ 23
(ii) 23 ÷ 28
(iii) (26)0
(iv) (30)6
(v) 83 × 8-5 × 84
(vi) 54 × 53 + 55
(vii) 54 ÷ 53 × 55
(viii) 44 ÷ 43 × 40
(ix) (35 × 47 × 58)

3. Simplify, giving solution with positive index :
(i) 2b6 ∙ b3 ∙ 5b4
(ii) x2y3 ∙ 6x5y ∙ 9x3y4
(iii) (-a5) (a2)
(iv) (-y2) (-y3)
(v) (-3)2 (3)3
(vi) (-4x) (-5x2)
(vii) (5a2b)(2ab2) (a3b)
(viii) x2a + 7 ∙ x2a – 8
(ix) 3y∙32 ∙ 3-4
(x) 24a∙33a ∙ 2-a
(xi) 4x2y2 ÷ 9x3y3
(xii) (102)3 (x8)12
(xiii) (a10)10 (16)10
(xiv) (n2)2 (-n2)2
(xv) -(3ab)2 (-5a2bc4)2
(xvi) (-2)2  × (0)3 × (3)3
(xvii) (2a3)4 (4a2)2
(xviii) (4x2y3)3 ÷ (3x2y3)
(xix) (1/2x)3 × (6x)2
(xx) (1/4ab2c)2 ÷ (3/2a2bc2)4
(xxi) [(5x7)3 ∙ (10x2)2]/(2x6)7
(xxii) [(7p2q9r5)2 (4pqr)3]/(14p6q10r4)2

4. Simplify and express the solution in the positive exponent form :
(i) [(-3)3 × 26 ]/(6 ×23)
(ii) [(23)5 × 54]/(43 × 52)
(iii) [36×(-6)2 ×36]/(123 × 35)
(iv) -128/2187
(v) [a-7 × b-7 × c5 × d4]/[a3 × b-5 × c-3 × d8]
(vi) (a3b-5)-2

5. Evaluate
(i) 6-2 ÷ (4-2 × 3-2)
(ii) [(5/6)2 × 9/4]÷[(-32/2) × (125/216)]
(iii) 53 × 32 + (17)0 × 73
(iv) 25 × 150 + (-3)3 – (2/7)-2
(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3
(vi) 5n ×25n-1 ÷ (5n-1 × 25n-1)