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) 3^{2} × 3^{4} × 3^{8}

Solution:

3^{2} × 3^{4} × 3^{8} = 3^{2} + 4 + 8

(ii) 6^{15} ÷ 6^{10}

Solution:

6^{15} ÷ 6^{10} = 6^{15} ^{ – } ^{10} = 6^{5}

(iii) a^{3} × a^{2}

Solution:

a^{3} × a^{2} = a^{3} ^{+} ^{2} = a^{5}

(iv) 7^{x} x 7^{2}

Solution:

7^{x} × 7^{2} = 7^{x} ^{+} ^{2}

(v) (5^{2})^{3} ÷ 5^{3}

Solution:

(5^{2})^{3} ÷ 5^{3} = 5^{2} ^{×} ^{3} ÷ 5^{3}

= 5^{6} ÷ 5^{3} = 5^{6} ^{–} ^{3} = 5^{3}

(vi) 2^{5} × 5^{5}

Solution:

2^{5} × 5^{5} = (2 × 5)^{5} = (10)^{5}

(vii) a^{4} × b^{4}

Solution:

a^{4} × b^{4} = (ab)^{4}

(viii) (3^{4})^{3}

Solution:

(3^{4})^{3} = 3^{4} ^{×} ^{3} = 3^{12}

(ix) (2^{20} ÷ 2^{15}) × 2^{3}

Solution:

(2^{20} ÷ 2^{15}) × 2^{3} = (2^{10} ^{–} ^{15}) × 2^{3}

= 2^{5} × 2^{3} = 2^{5} ^{+} ^{3} = 2^{8}

(x) 8^{t} ÷ 8^{2}

Solution:

8^{t} ÷ 8^{2} = 8^{t} ^{–} ^{2}

Question 2.

Simplify and express each of the following in exponential form:

(i)

Solution:

=

[∵ 4 = 2 × 2 = 2^{2} , 32 = 2 × 2 × 2 × 2 × 2 = 2^{5}]

= =

= 2^{5} ^{–} ^{5} x 3^{4} ^{–} ^{1} = 2^{0} × 3^{3} = 1 × 3^{3} = 3^{3}

(ii) [(5^{2})^{3} × 5^{4}] ÷ 5^{7}

Solution:

[(5^{2})^{3} × 5^{4}] ÷ 5^{7} = (5 ^{2 ×} ^{3} × 5^{4}) ÷ 5^{7}

=(5^{6} × 5^{4}) ÷ 5^{7} = 5^{6} ^{+} ^{4} ÷ 5^{7}

= 5^{10} ÷ 5^{7} = 5^{10} ^{–} ^{7} = 5^{3}

(iii) 25^{4} ÷ 5^{3}

Solution:

25^{4} ÷ 5^{3} = (5^{2})^{4} ÷ 5^{3}

= 5^{2} ^{×} ^{4} ÷ 5^{3}

= 5^{8} ÷ 5^{3} = 5^{8} ^{–} ^{3}

= 5^{5}

(iv)

Solution:

= = 3^{1} ^{–} ^{1} × 7^{2} ^{–} ^{1} × 11^{8} ^{–} ^{3}

= 3^{0} × 7^{1} × 11^{5}

= 1 × 7 × 11^{5}

= 7 × 11^{5}

= 7 × 11^{5}

(v)

Solution:

= =

3^{7} ^{–} ^{7} = 3^{0} = 1

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

Solution:

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

(vii) 2^{0} x 3^{0} x 4^{0}

Solution:

2^{0} × 3^{0} × 4^{0} = 1 × 1 × 1 = 3

(viii) (3^{0} + 2^{0}) × 5^{0}

Solution:

(3^{0} + 2^{0}) × 5^{0} = (1 + 1) × 1 = 2 × 1 = 2

(ix)

Solution:

= =

= 2^{8} ^{–} ^{6} × a^{5} ^{–} ^{3} = 2^{2} × a^{2} = (2a)^{2}

(x) x a^{8}

Solution:

× a^{8} = (a^{5} ^{–} ^{3}) × a^{8}

= a^{2} × a^{8}

= a^{2} ^{+} ^{8}

= a^{10}

(xi)

Solution:

= 4^{5} ^{–} ^{5} × a^{8} ^{–} ^{5} × b^{3} ^{–} ^{2}

= 4^{0} × a^{3} × b^{1}

= 1 × a × b

= a^{3}b

(xii) (2^{3} × 2)^{2}

Solution:

(2^{3} × 2)^{2} = (2^{3} + 1)^{2} = (2^{4})^{2} = 2^{8}

Question 3.

Say true or false and justify your answer:

(i) 10 × 10^{11} = 100^{11}

Solution:

(ii) 2^{3} > 5^{2}

Solution:

(iii) 2^{3} × 3^{2} = 6^{5}

Solution:

(iv) 3^{0} = (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)

= 2^{2} × 3^{3} × 2^{6} × 3^{1}

= 2^{2} ^{+} ^{6} × 3^{3} ^{+} ^{1}

= 2^{8} × 3^{4}

(ii) 270

Solution:

We have

∴ 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3^{3} × 5

(iii) 729 x 64

Solution:

We have

∴ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3^{2}

64 = 2 × 2 × 2 × 2 × 2 × 2 = 2^{6}

∴ 729 × 64 = 3^{6} × 2^{6}

(iv) 768

Solution:

We have

∴ 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2^{8} × 3

Question 5.

Simplify:

(i)

Solution:

= =

= 2^{10} ^{–} ^{9} × 7^{3} ^{–} ^{1}

= 2^{1} × 7^{2}

= 2 × 49

= 98

(ii)

Solution:

= = = = =

(iii)

Solution:

= =

= 3^{5} ^{–} ^{5} × 2^{5} ^{–} ^{5} × 5^{5} ^{+} ^{2} ^{–} ^{7
}= 30 × 20 × 50

= 1 × 1 × 1

= 1