RBSE Solutions for Class 7 Maths Chapter 5 Powers and Exponents Ex 5.2 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 5 Powers and Exponents Exercise 5.2.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 7 |
| Subject | Maths |
| Chapter | Chapter 5 |
| Chapter Name | Powers and Exponents |
| Exercise | Ex 5.2 |
| Number of Questions | 3 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 7 Maths Chapter 5 Powers and Exponents Ex 5.2
Question 1
Solve the following using (RBSESolutions.com) laws of exponents:
(i) 37 x 38
(ii) (4)7 x (4)2
(iii) a5 x a4
(iv) 315 ÷ 39
(v) t7 ÷ t4
(vi) (64 x 62) ÷ 65
(vii) (26)3
(viii) (a5)4
(ix) 55 x 85
(x) a3 x b3
(xi) 75 ÷ 65
(xii) (253 x 257) ÷ 2510
(xiii) 75 ÷ 78
(xiv) (93)0
Solution:
(i) 37 x 38 = (3)7+8 = (3)15 [ ∵ am x an = am+n ]
(ii) (4)7 x (4)2 = (4)7+2 = (4)9 [∵ am x n = am+n]
(iii) a5 x a4 = (a)5+4 = (a)9
(iv) 315 ÷ 39 = (3)15-9 = 36 [∵ am x an = am-n]
(v) t7 ÷ t4 = (t)7-4 = (t)3 [∵ am + an= am-n]
(vi) (64 x 62) ÷ 65 = (6)4+2 ÷ 65
= (6)6+ 65 = (6)6-5 = (6)1 = 6
(vii) (26)3 = (2)6×3 = 218 [∵ (am)n = amxn]
(viii) (a5)4 = (a)5×4= a2
(ix) 55 x 85 = (5 x 8)5 = (40)5
(x) a3 x b3 = (a + b)3 = (ab)3
(xi) 75 ÷ 65 = 75 x \(\frac { 1 }{ { 6 }^{ 5 } }\) = (\(\frac { 7 }{ 6 }\))5
(xii) (253 x 257) ÷ 2510
=(25)3+7 ÷ 2510
= (25)10 ÷ 2510
=(25)10-10= (25)0 = 1 [∵(a)0 = 1]
(xiii) 75 ÷ 78 = (7)5-8 = (7)-3= \(\frac { 1 }{ { 7 }^{ 3 } }\)
(xiv) (93)0 = (9)3×0 = (9)0 = 1
Question 2
Simplify the (RBSESolutions.com) following:
(i) {(32)3 x 34} ÷ 37
(ii) 164 ÷ 42
(iii) \(\frac { { 5 }^{ 7 } }{ { 5 }^{ 4 }\times { 5 }^{ 3 } }\)
(iv) 40 x 50 x 60
(v) \(\frac { { 3 }^{ 9 }\times { a }^{ 6 } }{ { 9 }^{ 2 }\times { a }^{ 3 } }\)
(vi) (73 x 7)3
(vii) \(\frac { { 3 }^{ 10 } }{ { 3 }^{ 5 }\times { 3 }^{ 7 } }\)
(viii) \(\frac { { a }^{ 9 } }{ { a }^{ 6 } }\)
(ix) 20 + 30 + 40
Solution:
(i) {(32)3 x 34 } ÷ 37
= {(3)2×3x 34 ÷ 37 = {(3)6 x 34 ) ÷ 37
= (3)6+4 ÷ (3)7
= (3)10 ÷ (3)7
= 310-7 = 33
(ii) 164 ÷ 42
= (4 x 4)4 ÷ 42
= (41+1)4 ÷ (4)2
= (4)8 ÷ (4)2 = (4)8-2 = (4)6
(iv) 40 x 50 x 60
= (4)0 x (5)0 x (6)0
= 1 x 1 x 1 = 1 {(a)0 = 1}
(vi) (73 x 7)3
= {(7)3+1}3 = (74)3 = (7)4 x 3= 712
(ix) 20 + 30 + 40
= (2)0 + (3)0 + (4)0 = 1 + 1 + 1 = 3
Question 3
Simplify the (RBSESolutions.com) following:
Solution:
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