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) 3^{7} x 3^{8}

(ii) (4)^{7} x (4)^{2}

(iii) a^{5} x a^{4}

(iv) 3^{15} ÷ 3^{9}

(v) t^{7} ÷ t^{4}

(vi) (6^{4} x 6^{2}) ÷ 6^{5}

(vii) (2^{6})^{3}

(viii) (a^{5})^{4}

(ix) 5^{5} x 8^{5}

(x) a^{3} x b^{3}

(xi) 7^{5} ÷ 6^{5}

(xii) (25^{3} x 25^{7}) ÷ 25^{10}

(xiii) 7^{5} ÷ 7^{8}

(xiv) (9^{3})^{0}

Solution:

(i) 3^{7} x 3^{8} = (3)^{7+8} = (3)^{15} [ ∵ a^{m} x a^{n} = a^{m+n} ]

(ii) (4)^{7} x (4)^{2} = (4)^{7+2} = (4)^{9} [∵ a^{m} x ^{n} = a^{m+n}]

(iii) a^{5} x a^{4} = (a)^{5+4} = (a)^{9}

(iv) 3^{15} ÷ 3^{9} = (3)^{15-9} = 3^{6} [∵ a^{m} x a^{n} = a^{m-n}]

(v) t^{7} ÷ t^{4} = (t)^{7-4} = (t)^{3} [∵ a^{m} + a^{n}= a^{m-n}]

(vi) (6^{4} x 6^{2}) ÷ 6^{5} = (6)^{4+2} ÷ 6^{5}

= (6)^{6}+ 6^{5} = (6)^{6-5} = (6)^{1} = 6

(vii) (2^{6})^{3} = (2)^{6×3} = 2^{18} [∵ (am)^{n} = a^{mxn}]

(viii) (a^{5})^{4} = (a)^{5×4}= a^{2}

(ix) 5^{5} x 8^{5} = (5 x 8)^{5} = (40)^{5}

(x) a^{3} x b^{3} = (a + b)^{3} = (ab)^{3}

(xi) 7^{5} ÷ 6^{5} = 7^{5} x [latex]\frac { 1 }{ { 6 }^{ 5 } }[/latex] = ([latex]\frac { 7 }{ 6 }[/latex])^{5}

(xii) (25^{3} x 25^{7}) ÷ 25^{10}

=(25)^{3+7} ÷ 25^{10}

= (25)^{10} ÷ 25^{10}

=(25)^{10-10}= (25)^{0} = 1 [∵(a)^{0} = 1]

(xiii) 7^{5} ÷ 7^{8} = (7)^{5-8} = (7)^{-3}= [latex]\frac { 1 }{ { 7 }^{ 3 } }[/latex]

(xiv) (9^{3})^{0} = (9)^{3×0} = (9)^{0} = 1

Question 2

Simplify the (RBSESolutions.com) following:

(i) {(3^{2})^{3} x 3^{4}} ÷ 3^{7}

(ii) 16^{4} ÷ 4^{2}

(iii) [latex]\frac { { 5 }^{ 7 } }{ { 5 }^{ 4 }\times { 5 }^{ 3 } }[/latex]

(iv) 4^{0} x 5^{0} x 6^{0}

(v) [latex]\frac { { 3 }^{ 9 }\times { a }^{ 6 } }{ { 9 }^{ 2 }\times { a }^{ 3 } }[/latex]

(vi) (7^{3} x 7)^{3}

(vii) [latex]\frac { { 3 }^{ 10 } }{ { 3 }^{ 5 }\times { 3 }^{ 7 } }[/latex]

(viii) [latex]\frac { { a }^{ 9 } }{ { a }^{ 6 } }[/latex]

(ix) 2^{0} + 3^{0} + 4^{0}

Solution:

(i) {(3^{2})^{3} x 3^{4} } ÷ 3^{7}

= {(3)^{2×3}x 3^{4} ÷ 3^{7} = {(3)^{6} x 3^{4} ) ÷ 3^{7}

= (3)^{6+4} ÷ (3)^{7}

= (3)^{10} ÷ (3)^{7}

= 3^{10-7} = 3^{3}

(ii) 16^{4} ÷ 4^{2}

= (4 x 4)^{4} ÷ 4^{2}

= (4^{1+1})^{4} ÷ (4)^{2}

= (4)^{8} ÷ (4)^{2} = (4)^{8-2} = (4)^{6}

(iv) 4^{0} x 5^{0} x 6^{0}

= (4)^{0} x (5)^{0} x (6)^{0}

= 1 x 1 x 1 = 1 {(a)^{0} = 1}

(vi) (7^{3} x 7)^{3}

= {(7)^{3+1}}^{3} = (7^{4})^{3} = (7)^{4 x 3}= 7^{12}

(ix) 2^{0} + 3^{0} + 4^{0}

= (2)^{0} + (3)^{0} + (4)^{0} = 1 + 1 + 1 = 3

Question 3

Simplify the (RBSESolutions.com) following:

Solution:

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