CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1 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 13.1

Question 1.

Find the value of:

(i) 2^{6}

Solution:

2^{6} = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9^{3}

Solution:

9^{2} = 9 × 9 × 9 = 729

(iii) 11^{2}

Solution:

11^{2} = 11 × 11 = 121

(iv) 5^{4}

Solution:

5^{4} = 5 × 5 × 5 × 5 = 625

Question 2.

Express the following in exponential form :

(i) 6 × 6 × 6 × 6

Solution:

6 × 6 × 6 × 6 = 6^{4}

(ii) t x t

Solution:

t × t = t^{2}

(iii) b × b × b × b

Solution:

b × b × b × b = b^{4}

(iv) 5 × 5 × 7 × 7 × 7

Solution:

5 × 5 × 7 × 7 × 7 = 5^{2} × 7^{3}

(v) 2 × 2 × a × a

Solution:

2 × 2 × a × a = 2^{2} × a^{2}

(vi) a × a × a × c × c × c × c × d

Solution:

a × a × a × c × c × c × c d = a^{3} × c^{4} × d

Question 3.

Express each of the following numbers using the exponential notation:

(i) 512

Solution:

We have

(ii) 343

Solution:

We have

(iii) 729

Solution:

We have

(vi) 3125

Solution:

We have

Question 4.

Identify the greater number, wherever possible, in each of the following?

(i) 4^{3} or 3^{4}

Solution:

(ii) 5^{3} or 3^{5}

Solution:

(iii) 2^{8} or 8^{2}

Solution:

(iv) 100^{2} or 2^{100}

Solution:

(v) 2^{10} or 10^{2}

Solution:

Question 5.

Express each of the following as a product of powers of their prime factors:

(i) 648

Solution:

To determine the prime factorization of 648, we use the division method, as shown below:

(ii) 405

Solution:

We use the division method as shown under:

(iii) 540

Solution:

We use the division method as shown under:

(iv) 3600

Solution:

We use the division method as shown under:

Question 6.

Simplify:

(i) 2 × 10^{3}

Solution:

2 × 10^{3} = 2 × 1000 = 2000

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

Solution:

7^{2} × 2^{2} = (7 × 2)^{2} = (14)^{3} = 14 × 14 = 196

(iii) 2^{3} × 5

Solution:

2^{3} × 5 = 2 × 2 × 2 × 5

= 8 × 5 = 40

(iv) 3 × 4^{4}

Solution:

3 × 4^{4} = 3 × 4 × 4 × 4 × 4

= 3 × 256 = 768

(v) 0 × 10^{2}

Solution:

0 × 10^{2} = 0

(vi) 5^{2} × 3^{3}

Solution:

5^{2} × 3^{3} = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675

(vii) 24 × 32

Solution:

2^{4} × 3^{2} = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144

(viii) 3^{2} × 10^{4}

Solution:

3^{2} × 10^{4} = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

Question 7.

Simplify:

(i) (-4)^{3}

Solution:

(- 4)^{3} = (- 4) × (- 4) × (- 4) = – 64

(ii) (-3) × (-2)^{3}

Solution:

(-3) × (-2)^{3} = (-3) × (- 2) × (- 2) × (- 2) = (- 3) × (- 8) = 24

(iii) (- 3)^{2} × (- 5)2

Solution:

(- 3)^{2} × (- 5)^{2} = (- 3) × (- 3) × (- 5) × (- 5) = 9 × 25 = 225

(iv) (-2)^{3} × (-10)^{2}

Solution:

(-2)^{3} × (-10)^{3} = (-2) × (-2) × (-2) x (-10) × (-10) × (-10) = (-8) × (-1000) = 8000

Question 8.

Compare the following numbers:

(i) 2.7 × 10^{12 }; 1.5 × 10^{8}

Solution:

We have,

2.7 × 10^{12} = 2.7 × 10 × 10^{11}

= 27 × 10^{11}, contains 13 digits and 1.5 × 10^{8} = 1.5 × 10 × 10^{7}

= 15 × 10^{7} contains 9 digits

Clearly, 2.7 × 10^{12} > 1.5 × 10^{8}

(ii) 4 × 10^{14}; 3 × 10^{17}

Solution:

We have, 4 × 10^{14}, contains 15 digits and, 3 × 10^{17}, contains 18 digits 4 × 10^{14} < 3 × 10^{17}