# NCERT Solutions for Class 7 Maths Chapter 13 Exercise 13.2

NCERT Solutions for Class 7 Maths Chapter 13 Exercise 13.2 (Ex. 13.2) Exponents and Powers in Hindi and English Medium for CBSE Session 2021-2022. All the solutions and study material is updated according to latest NCERT Textbooks.

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## Class 7 Maths Chapter 13 Exercise 13.2 Solution

• ### CBSE NCERT Class 7 Maths Chapter 13 Exercise 13.2 Solution in Hindi and English Medium

 Class: 7 Mathematics Chapter: 13 Exponents and Powers Exercise: 13.2 Hindi and English Medium Solution

### Class 7 Maths Chapter 13 Exercise 13.2 Solution in Videos

#### Power of a Power

Consider the following
Simplify {(2)³}² = (2)³ x (2)³
Or, (2)³⁺³ = (2)⁶

##### Multiplying Powers with the Same Exponents

simplify 2³ × 3³? Notice that here the two terms 2³ and 3³ have different bases, but the same exponents.
Now, 2³ × 3³ = (2 x 2 x 2) x (3 x 3 x 3)
Or, (2 x 3) x (2 x 3) x (2 x 3)
= 6 x 6 x 6 = 6³ (Observe 6 is the product of bases 2 and 3)

In general, for any non-zero integer a
Aᵐ × bᵐ = (ab)ᵐ (where m is any whole number)

### Class 7 Maths Exercise 13.2 Important Questions

##### Simplify and write the answer in exponential form: (i) (62)4 (ii) (22)100 (iii) (750)2 (iv) (53)7

We have:
(i) (62)4 = 62 x 62 x 62 x 62 = 62+2+2+2 = 62×4 = 68
(ii) (22)100 = 22×100 = 2200
(iii) (750)2 = 750×2 = 7100
(iv) (53)7 = 53×7 = 521

##### Express the following terms in the exponential form: (i) (2 × 3)⁵ (ii) (2a)⁴ (iii) (– 4m)³

We have:
(i) (2 × 3)⁵ = (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3)
= (2 × 2 × 2 × 2 × 2) × (3 × 3× 3 × 3 × 3)
= 2⁵ × 3⁵
(ii) (2a)⁴ = 2a × 2a × 2a × 2a = (2 × 2 × 2 × 2) × (a × a × a × a) = 2⁴ × a⁴
(iii) (– 4m)³ = (– 4 × m)³ = (– 4 × m) × (– 4 × m) × (– 4 × m) = (– 4) × (– 4) × (– 4) × (m × m × m)
= (– 4)³ × (m)³

### Class 7 Maths Exercise 13.2 Important Questions

##### Simplify: (6⁻¹ – 8⁻¹)⁻¹ + (2⁻¹ – 3⁻¹)⁻¹

We have:
(6⁻¹ – 8⁻¹)⁻¹ + (2⁻¹ – 3⁻¹)⁻¹
= (1/6 – 1/8)⁻¹ + (1/2 – 1/3)⁻¹
= {(4 – 3)/24}-¹ + {(3 – 2)/6}-¹
= (1/24)⁻¹ + (1/6)⁻¹
= (24/1) + (6/1)
= 24 + 6 = 30

##### By what number should we multiply 3⁻⁹ so that the product is equal to 3?

Let the required number be x.
Then, 3⁻⁹ X x = 3
Or, x = 3/3⁻⁹
Or, x = 3 X 3⁹
X = 3¹⁺⁹ = 3¹⁰
Hence, the required number is 3¹⁰

##### By what number should we multiply (-8)⁻¹ to obtain a product equal to 10⁻¹?

Let the required number be x. Then,
Then, (-8)⁻¹ X x = 10⁻¹
Or, x = 10⁻¹/(-8)⁻¹
Or, x = -8 / 10
X = -4/5        