RBSE Solutions for Class 8 Maths Chapter 3 Powers and Exponents Ex 3.1 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 3 Powers and Exponents Exercise 3.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Powers and Exponents |

Exercise |
Exercise 3.1 |

Number of Questions |
10 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 3 Powers and Exponents Ex 3.1

Question 1.

Simplify the following

Solution.

Question 2.

Find the values

Solution.

(i) (-5)^{3}

= (-1 × 5)^{3}

= (-1)^{3} × (5)^{3}

= (-1) × 5 × 5 × 5

= – 125

Question 3.

With the help of Prime factor, change the following into exponent form

Solution.

Question 4.

Find the values

Solution.

(i) 3^{2} × 3^{3}

3^{2+3} = 3^{5} = 3 × 3 × 3 × 3 × 3

= 243

Question 5.

Answer in exponent form-

Solution.

(i) 4^{5} ÷ 4^{2}

= 4^{5} ÷ 4^{2}

= 4^{5–}^{2}

= 4^{3}

(ii) (- 5)^{7} ÷ (- 5)^{4}

= (- 5)^{7-4}

=(- 5)^{3}

Question 6.

Find the value

(i) (3^{2})^{3}

(ii) (2^{3})^{2}

(iii) (5^{2})^{2}

(iv) (-2^{4})^{2}

Solution.

(i) (3^{2})^{3}

= (3 x 3)^{3}

=9^{3}

= 9 x 9 x 9 = 729

(ii) (2^{3})^{2}

= 2^{3} x 2

= 2^{6}

= 2 x 2 x 2 x 2 x 2 x 2=64

(iii) (5^{2})^{2
}= 5^{2×2
}= 5^{4}

= 5 x 5 x 5 x 5 = 625

(iv) (- 2^{4})^{2}

= (- 2 x 2 x 2 x 2)^{2}

= (- 16)^{2}

= {(- 1) x 16)^{2}

= (- 1)^{2} x (16)^{2}

= 1 x 16 x 16 = 256

Question 7.

Find the value

(i) 3^{0}

(ii) 7^{5-5}

(iii) (-2)^{3-3}

(iv) \({ \left( \frac { 2 }{ 3 } \right) }^{ 2+3-5 }\)

(v) 2^{0} × 3^{0}

(vi) 2^{0} + 5^{0}

(vii) \({ \left( \frac { 7 }{ 15 } \right) }^{ 0 }+{ \left( \frac { 1 }{ 7 } \right) }^{ 3-3 }\)

Solution.

(i) 3^{0}

3^{0} = 1

(ii) 7^{5-5}

7^{5-5} = 7^{0 }= 1

(iii) (- 2)^{3-3}

(- 2)^{3-3} = (- 2)^{0} = 1

(iv) \({ \left( \frac { 2 }{ 3 } \right) }^{ 2+3-5 }\)

\({ \left( \frac { 2 }{ 5 } \right) }^{ 2+3-5 }\)

\({ \left( \frac { 2 }{ 5 } \right) }^{ 0 }=1\)

(v) 2^{0} X 3^{0}

2^{0} x 3^{0} = 1 x 1 = 1

(vi)2^{0 }+ 5^{0}

2^{0} + 5^{0} = 1 + 1 = 2

Question 8.

Change into positive exponent numbers

(i) 2^{-3}

(ii) 3^{-5}

(iii) a^{-4}

(iv) (-2)^{-5}

(v) (-x)^{-3}

(vi) \(\frac { 1 }{ { 5 }^{ -3 } } \)

(vii) \(\frac { 1 }{ { y }^{ -3 } } \)

(viii) \(\frac { 1 }{ { \left( \frac { 2 }{ 3 } \right) }^{ -3 } } \)

Solution.

(i) 2^{-3}

Question 9.

Simplify the following ¡n form of exponent

Solution.

(i) (2^{2} x 3^{3})^{2}

= (2 x 2 x 3 x 3 x 3)^{2}

= (108)^{2}

= 108 X 108

= 11664

Question 10.

Find the values of

Solution.

= (3)² + (2)² + (4)²

=9 + 4 + 16

= 29

We hope the given RBSE Solutions for Class 8 Maths Chapter 3 Powers and Exponents Ex 3.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 3 Powers and Exponents Exercise 3.1, drop a comment below and we will get back to you at the earliest.

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