Ex-6.2, Exponents, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics PDF Download

Q1. Using laws of exponents, simplify and write the answer in exponential form

(i) 2³×2⁴×2⁵

(ii) 5¹²÷5³

(iii) (7²)³

(iv) (3²)⁵÷34

(v) 3⁷×2⁷

(vi) (5²¹÷5¹³)×5⁷

Sol:

(i) 2³×2⁴×2⁵

We know that, a^m+aⁿ+a^p = a^m+n+p

So, 2³×2⁴×2⁵ = 2³⁺⁴⁺⁵

= 2¹²

(ii) 5¹²÷5³

We know that, a^m÷aⁿ = a^m−n

So, 5¹²÷5³ = 5¹²⁻³

= 5⁹

(iii) (7²)³

We know that, (am)n=amn

So, (72)3 = 7(2)(3)

= 76

(iv) (3²)⁵÷3⁴

We know that, a^m÷aⁿ = a^m−n and (a^m)ⁿ = a^mn

So, (3²)⁵÷3⁴ = 3¹⁰÷3⁴

= 3¹⁰⁻⁴

= 3⁶

(v) 3⁷×2⁷

We know that, (a^m×b^m)=(a×b)^m

So, 3⁷×2⁷ = (3×2)⁷

= 6⁷

(vi) (5²¹÷5¹³)×5⁷

We know that, a^m÷aⁿ=a^m−n and(a^m×aⁿ)=(a)^m+n

So, (5²¹÷5¹³)×5⁷ = (5²¹⁻¹³)×5⁷

= (5⁸)×5⁷

= 5⁸⁺⁷

= 5¹⁵

Q2. Simplify and express each of the following in exponential form

Sol:

Q3. Simplify and express each of the following in exponential form

Sol:

We know that,

We know that,

Q4. Write 9 ×9 ×9 ×9 ×9 in exponential form with base 3

Sol:

9 ×9 ×9 ×9 ×9 = (9)⁵ = (3²)⁵

= 3¹⁰

Q5. Simplify and write each of the following in exponential form

Sol:

Q6. Simplify

Sol:

Q7. Find the values of n in each of the following

Sol:

Equating the powers

= 2n + 3 = 11

= 2n = 11- 3

=  2n = 8

= n = 4

(ii) 9×3ⁿ = 3⁷

= 3²×3ⁿ = 3⁷

= 3²⁺ⁿ = 3⁷

Equating the powers

= 2 + n = 7

= n = 7 – 2

= n = 5

Equating the powers

= n + 5 = 5

= n = 0

Equating the powers

Equating the powers

= 4 + 5 = 2n + 1

= 2n + 1 = 9

= 2n = 8

= n = 4

= 3(2n – 2) =2(2n – 2)

= 6n – 6 = 4n – 4

= 6n – 4n = 6 – 4

= 2n = 2

= n = 1

Q8.  find the value of n

Sol:

On equating the coefficient

3n – 15 = -3

3n = -3 + 15

3n = 12

n = 4