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 2022-2023. All the solutions and study material is updated according to latest NCERT Textbooks. If student feel any difficulty in PDF solutions, they may refer to video solution to understand properly. Videos related to each exercises are given separately on website as well as in app.

## 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