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:
(ii) 23 > 52
Solution:
(iii) 23 × 32 = 65
Solution:
(iv) 30 = (1000)0
Solution:
Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
Solution:
We have
∴ 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
∴ 270 = 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5
(iii) 729 x 64
Solution:
We have
∴ 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
∴ 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