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NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2

CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 Textbook Questions and Answers are provided by experts in order to help students secure good marks in exams.

Class 7 Maths NCERT Solutions Chapter 13 Exponents and Powers Ex 13.2

Question 1.
Using the law of exponents, simplify and write the answer in exponential form:

(i) 32 × 34 × 38
Solution:
32 × 34 × 38 = 32 + 4 + 8

(ii) 615 ÷ 610
Solution:
615 ÷ 610 = 615 – 10 = 65

(iii) a3 × a2
Solution:
a3 × a2 = a3 + 2 = a5

(iv) 7x x 72
Solution:
7x × 72 = 7x + 2

(v) (52)3 ÷ 53
Solution:
(52)3 ÷ 53 = 52 × 3 ÷ 53
= 56 ÷ 53 = 56 – 3 = 53

(vi) 25 × 55
Solution:
25 × 55 = (2 × 5)5 = (10)5

(vii) a4 × b4
Solution:
a4 × b4 = (ab)4

(viii) (34)3
Solution:
(34)3 = 34 × 3 = 312

(ix) (220 ÷ 215) × 23
Solution:
(220 ÷ 215) × 23 = (210 – 15) × 23
= 25 × 23 = 25 + 3 = 28

(x) 8t ÷ 82
Solution:
8t ÷ 82 = 8t – 2

Question 2.
Simplify and express each of the following in exponential form:
(i) \frac{2^{3} \times 3^{4} \times 4}{3 \times 32}
Solution:
\frac{2^{3} \times 3^{4} \times 4}{3 \times 32} = \frac{2^{3} \times 3^{4} \times 2^{2}}{3 \times 2^{5}}
[∵ 4 = 2 × 2 = 22 , 32 = 2 × 2 × 2 × 2 × 2 = 25]
= \frac{2^{3+2} \times 3^{4}}{3^{1} \times 2^{5}} = \frac{2^{5} \times 3^{4}}{3^{1} \times 2^{5}}
= 25 – 5 x 34 – 1 = 20 × 33 = 1 × 33 = 33

(ii) [(52)3 × 54] ÷ 57
Solution:
[(52)3 × 54] ÷ 57 = (5 2 × 3 × 54) ÷ 57
=(56 × 54) ÷ 57 = 56 + 4 ÷ 57
= 510 ÷ 57 = 510 – 7 = 53

(iii) 254 ÷ 53
Solution:
254 ÷ 53 = (52)4 ÷ 53
= 52 × 4 ÷ 53
= 58 ÷ 53 = 58 – 3
= 55

(iv) \frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}
Solution:
\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}} = \frac{3 \times 7^{2} \times 11^{8}}{3 \times 7 \times 11^{3}} = 31 – 1 × 72 – 1 × 118 – 3
= 30 × 71 × 115
= 1 × 7 × 115
= 7 × 115
= 7 × 115

(v) \frac{3^{7}}{3^{4} \times 3^{3}}
Solution:
\frac{3^{7}}{3^{4} \times 3^{3}} = \frac{3^{7}}{3^{4+3}} = \frac{3^{7}}{3^{7}}
37 – 7 = 30 = 1

(vi) 20 + 30 + 40
Solution:
20 + 30 + 40 = 1 + 1 + 1 = 3

(vii) 20 x 30 x 40
Solution:
20 × 30 × 40 = 1 × 1 × 1 = 3

(viii) (30 + 20) × 50
Solution:
(30 + 20) × 50 = (1 + 1) × 1 = 2 × 1 = 2

(ix) \frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}
Solution:
\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}} = \frac{2^{8} \times a^{5}}{\left(2^{2}\right)^{3} \times a^{3}} = \frac{2^{8} \times a^{5}}{2^{6} \times a^{3}}
= 28 – 6 × a5 – 3 = 22 × a2 = (2a)2

(x) \left(\frac{a^{5}}{a^{3}}\right) x a8
Solution:
\left(\frac{a^{5}}{a^{3}}\right) × a8 = (a5 – 3) × a8
= a2 × a8
= a2 + 8
= a10

(xi) \frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}
Solution:
\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}} = 45 – 5 × a8 – 5 × b3 – 2
= 40 × a3 × b1
= 1 × a × b
= a3b

(xii) (23 × 2)2
Solution:
(23 × 2)2 = (23 + 1)2 = (24)2 = 28

Question 3.
Say true or false and justify your answer:

(i) 10 × 1011 = 10011
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 58

(ii) 23 > 52
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 55

(iii) 23 × 32 = 65
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 56

(iv) 30 = (1000)0
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 57

Question 4.
Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 50
∴ 108 × 192 = (2 × 2 × 3 × 3 × 3) x (2 × 2 × 2 × 2 × 2 × 2 × 3)
= 22 × 33 × 26 × 31
= 22 + 6 × 33 + 1
= 28 × 34

(ii) 270
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 51
∴ 270 = 2 × 3 × 3 ×  3 × 5 = 2 × 33 × 5

(iii) 729 x 64
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 52
∴ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 32
64 = 2 × 2 × 2 × 2 × 2 × 2 = 26
∴ 729 × 64 = 36 × 26

(iv) 768
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 53
∴ 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 3

Question 5.
Simplify:

(i) \frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}
Solution:
\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7} = \frac{2^{10} \times 7^{3}}{\left(2^{3}\right)^{3} \times 7} = \frac{2^{10} \times 7^{3}}{2^{9} \times 7}
= 210 – 9  × 73 – 1
= 21 × 72
= 2 × 49
= 98

(ii) \frac{\mathbf{2 5} \times \mathbf{5}^{2} \times t^{8}}{10^{3} \times t^{4}}
Solution:
\frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}} = \frac{5^{2} \times 5^{2} \times t^{8}}{(2 \times 5)^{3} \times t^{4}} = \frac{5^{4} \times t^{8}}{2^{3} \times 5^{3} \times t^{4}} = \frac{5^{4-3} \times t^{8-4}}{2^{3}} = \frac{5 \times t^{4}}{2^{3}} = \frac{5 t^{4}}{8}

(iii) \frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}
Solution:
\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}} = \frac{3^{5} \times(2 \times 5)^{5} \times 5^{2}}{5^{7} \times(2 \times 3)^{5}} = \frac{3^{5} \times 2^{5} \times 5^{5} \times 5^{2}}{5^{7} \times 2^{5} \times 3^{5}}
= 35 – 5 × 25 – 5 × 55 + 2 – 7
= 30 × 20 × 50
= 1 × 1 × 1
= 1