NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

Class 7 Maths NCERT Solutions Chapter 13 Exponents and Powers Ex 13.1

Question 1.
Find the value of:

(i) 2⁶
Solution:
2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9³
Solution:
9² = 9 × 9 × 9 = 729

(iii) 11²
Solution:
11² = 11 × 11 = 121

(iv) 5⁴
Solution:
5⁴ = 5 × 5 × 5 × 5 = 625

Question 2.
Express the following in exponential form :

(i) 6 × 6 × 6 × 6
Solution:
6 × 6 × 6 × 6 = 6⁴

(ii) t x t
Solution:
t × t = t²

(iii) b × b × b × b
Solution:
b × b × b × b = b⁴

(iv) 5 × 5 × 7 × 7 × 7
Solution:
5 × 5 × 7 × 7 × 7 = 5² × 7³

(v) 2 × 2 × a × a
Solution:
2 × 2 × a × a = 2² × a²

(vi) a × a × a × c × c × c × c × d
Solution:
a × a × a × c × c × c × c  d = a³ × c⁴ × 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) 4³ or 3⁴
Solution:

(ii) 5³ or 3⁵
Solution:

(iii) 2⁸ or 8²
Solution:

(iv) 100² or 2¹⁰⁰
Solution:

(v) 2¹⁰ or 10²
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 × 10³
Solution:
2 × 10³ = 2 × 1000 = 2000

(ii) 7² x 2²
Solution:
7² × 2² = (7 × 2)² = (14)³ = 14 × 14 = 196

(iii) 2³ × 5
Solution:
2³ × 5 = 2 × 2 × 2 × 5
= 8 × 5 = 40

(iv) 3 × 4⁴
Solution:
3 × 4⁴ = 3 × 4 × 4 × 4 × 4
= 3 × 256 = 768

(v) 0 × 10²
Solution:
0 × 10² = 0

(vi) 5² × 3³
Solution:
5² × 3³ = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675

(vii) 24 × 32
Solution:
2⁴ × 3² = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144

(viii) 3² × 10⁴
Solution:
3² × 10⁴ = 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 × 10¹² ; 1.5 × 10⁸
Solution:
We have,
2.7 × 10¹² = 2.7 × 10 × 10¹¹
= 27 × 10¹¹, contains 13 digits and 1.5 × 10⁸ = 1.5 × 10 × 10⁷
= 15 × 10⁷ contains 9 digits
Clearly, 2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴; 3 × 10¹⁷
Solution:
We have, 4 × 10¹⁴, contains 15 digits and, 3 × 10¹⁷, contains 18 digits 4 × 10¹⁴ < 3 × 10¹⁷