Selina Concise Solutions for Chapter 5 Exponents (including Laws of Exponents) Class 7 ICSE Mathematics
Exercise 5 (A)
1. Find the value of:
(i) 62
(ii) 73
(iii) 44
(iv) 55
(v) 83
(vi) 75
Answer
(i) 62 = 6 × 6 = 36
(ii) 73 = 7 × 7 × 7 = 343
(iii) 44 = 4 × 4 × 4 × 4 = 256
(iv) 55 = 5 × 5 × 5 × 5 × 5 = 3125
(v) 83 = 8 × 8 × 8 = 512
(vi) 75 = 7 × 7 × 7 × 7 × 7 =16807
2.Evaluate:
(i) 23 × 42
(ii) 23 × 52
(iii) 33 × 52
(iv) 22 × 33
(v) 32 × 52
(vi) 53 × 24
(vii) 32 × 42
(viii) (4 × 3)3
(ix) (5 × 4)2
Answer
(i) 23 × 42
= 2 × 2 × 2 × 4 × 4
= 8 × 16
= 128
(ii) 23 × 52
= 2 × 2 × 2 × 5 × 5
= 8 × 25
= 200
(iii) 33 × 52
= 3 × 3 × 3 × 5 × 5
= 27 × 25
= 675
(iv) 22 × 33
= 2 × 2 × 3 × 3 × 3
= 4 × 27
= 108
(v) 32 × 52
=3 × 3 × 5 × 5 × 5
= 9 × 125
= 1125
(vi) 53 × 24
= 5 × 5 × 5 × 2 × 2 × 2 × 2
= 125 × 16
= 2000
(vii) 32 × 42
=3 × 3 × 4 × 4
= 9 × 16
=144
(viii) (4 × 3)3
=4 × 4 × 4 × 3 × 3 × 3
= 64 × 27
= 1728
(ix) (5 × 4)2
=5 × 5 × 4 × 4
= 25 × 16
= 400
3. Evaluate :
(i) (3/4)4
(ii) (-5/6)5
(iii) (-3/-5)3
Answer
4. Evaluate :
(i) (2/3)3 × (3/4)2
(ii) (-3/4)3 × (2/3)4
(iii) (3/5)2 × (-2/3)3
Answer
5. Which is greater :
(i) 23 or 32
(ii) 25 or 52
(iii) 43 or 34
(iv) 54 or 45
Answer
(i) 23 or 32
Since, 23 = 2 × 2 × 2 = 8
and, 32 = 3 × 3 = 9
∵ 9 is greater than 8
⇒ 32 > 23
(ii) 25 or 52
Since, 25 = 2 × 2 × 2 × 2 × 2 = 32
and, 52 = 5 × 5 = 25
∵ 32 is greater than 25
⇒ 25 > 52
(iii) 43 or 34
Since, 43 = 4 × 4 × 4 = 64
and, 34 = 3 × 3 × 3 × 3 = 81
∵ 81 is greater than 64
⇒ 34 > 43
(iv) 54 or 45
Since, 54 = 5 × 5 × 5 × 5 = 625
and, 45 = 4 × 4 × 4 × 4 × 4= 1024
∵ 1024 is greater than 625
⇒ 45 > 54
6. Express each of the following in exponential form :
(i) 512
(ii) 1250
(iii) 1458
(iv) 3600
(v) 1350
(vi) 1176
Answer
⇒ 27 = 3x
⇒ 33 = 3x
Also, 32 = 2y
(i) the values of a and b.
(ii) 2b × 5a
64 × 625 = 2a × 5b
⇒ 64 = 2a
⇒ 64 = 26
∴ a = 6
Also, 625 = 5b
= 24 × 56
= 2×2×2×2×5×5×5×5×5×5
= 16 × 15625
Exercise 5 (B)
(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)0
(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)3
(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 2
(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
(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)
(ii) mn + nm
(iii) 6m-3 + 4n2
(iv) 2n3 – 3m
= (-2)2 + (2)2 - 2(-2)(2)
= 4 + 4 -(-8)