RD Sharma Solutions Class 7 Exponents Exercise 6.2

RD Sharma Class 7 Solutions Chapter 6 Ex 6.2 PDF Free Download

RD Sharma Solutions Class 7 Chapter 6 Exercise 6.2

Exercise 6.2

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

(i) 2× 2× 25

Solution: 23+4+5 = 212 [Since, am × an = am+n]

(ii) 512 ÷ 53

Solution: 512-3 = 59 [Since, a÷ an = am-n]

(iii) (72)3

Solution: 72×3 = 76 [Since, (am)n = am×n]

(iv) (32)5 ÷ 34

Solution: 32×5 ÷ 34 [Since, (am)n = am×n]

= 310 ÷ 34

= 310 – 4 [Since, a÷ an = am-n]

= 36

(v) 37 × 27

Solution: (3 × 2)= 6[Since, am × am = (a × b)m]

(vi) (521 ÷ 513) × 57

Solution: (521-13× 57 [Since, a÷ an = am-n]

= 58 × 57

= 515 [Since, am × an = am+n]

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

(i) {(23)4 × 28} ÷ 212

Solution: {24 × 28} ÷ 212

= {212 × 28} ÷ 212

= {212+8} ÷ 212

= {220} ÷ 212

= 220-12

= 28

(ii) (8× 84) ÷ 83

Solution: (82+4) ÷ 83

= (86) ÷ 83

= 86-3

= 83

= (23)3

= (2)9

(iii) RD Sharma Solutions Class 7 Maths Qs 2 (iii)

Solution: 57-2× 53

= 55 × 53

= 55+3

= 58

(iv) RD Sharma Solutions Class 7 Maths Qs 2 (iv)

Solution: 54-4 × x10-7 × y5-4

= 50 × x3 × y2

= 1 × x3 × y2

= x3 y2

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

(i) {(32)3 × 26} ÷ 56

Solution: {(33 × 26} ÷ 56

= {(36 × 26} ÷ 56

= {(66} ÷ 56

= 66 ÷ 56

= 306

(ii) RD Sharma Solutions Class 7 Maths Qs 3 (ii)

Solutions: RD Sharma Solutions Class 7 Maths Ans 3 (iv)

= x12 × y24-12 × 212

= x12 × y12 × 212

= (2xy)12

(iii) RD Sharma Solutions Class 7 Maths Qs 3 (iii)

Solution: RD Sharma Solutions Class 7 Maths Ans 3 (iii)

RD Sharma Solutions Class 7 Maths Sol 3 (iii)

(iv) RD Sharma Solutions Class 7 Maths Qs 3 (iv)

Solution: RD Sharma Solutions Class 7 Maths Ans 3 (iv)

RD Sharma Solutions Class 7 Maths Sol 3 (iv)

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

Solution: 95

= (32)5

= 310

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

(i) (25)3 ÷ 53

Solution: (52)3 ÷ 53

= 56 ÷ 53

= 56-3

= 53

(ii) (81)5 ÷ (32)5

Solution: (34)5 ÷ (32)5

= 320 ÷ 310

= 320-10

= 310

(iii) RD Sharma Solutions Class 7 Maths Qs 5 (iii)

Solution: RD Sharma Solutions Class 7 Maths Ans 5 (iii)

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Sol 5 (iii)

= 316-12 × x10-6

= 34 × x4

= (3x)4

(iv) RD Sharma Solutions Class 7 Maths Qs 5 (iv)

Solution: RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Ans 5 (iv)

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Sol 5 (iv)

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 5 (iv)

= 73 × 133

= 913

Q6. Simplify

(i) (35)11 × (315)4 – (35)18 × (35)5

Solution: 355 × 360 – 390 × 325

= 355+60 – 390+25

= 3115 – 3115

= 3115-115

= 30

= 1

(ii)

  RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Qs 6 (ii)

= 7/14

= 1/2

(iii) 

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Qs 6 (iii)

(iv)

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

(i) 52n × 53 = 511

Solution: 52n+3 = 511

Now, equating the coefficients

⇒ 2n + 3 = 11

⇒ 2n = 11 – 3

⇒ 2n = 8

⇒ n = 4

(ii) 9 × 3n = 37

Solution: 9 = 37-n

⇒ 32 = 37-n

Now, equating the coefficients

⇒ 2 = 7 – n

⇒ n = 7 – 2

⇒ n = 5

(iii) 8 × 2n+2 =32

Solution: 23 × 2n+2 = 25

⇒ 2n+2+3 = 25

⇒ 2n+5 = 25

Now, equating the coefficients

⇒ n + 5 = 5

⇒ n = 0

(iv) 72n+1 ÷ 49 = 73

Solution: 72n+1 ÷ 72 = 73

⇒  72n+1-2 = 73

⇒ 72n-1 = 73

Now, equating the coefficients

⇒ 2n – 1 = 3

⇒ 2n = 4

⇒ n = 2

(v) 

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Qs 7 (iii)

Equating the coefficients

⇒ 4 + 5 = 2n + 1

⇒ 9 = 2n + 1

⇒ 8 = 2n

⇒ n = 4

(vi)

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Qs 7 (vi)

Equating the coefficients

⇒ 3(2n – 2) = 2 (2n – 2)

⇒ 6n – 6 = 4n – 4

⇒ 6n – 4n = 6 – 4

⇒ 2n = 2

⇒ n = 1

Q8. If

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Qs 8

Find the value of n.

Solution: 

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Ans 8

RD Sharma Solutions Class 7 Maths Chapter 6 Exercise 6.2 Sol 8

On equating the coefficient

3n – 15 = -3

3n = -3 + 15

3n = 12

n = 4

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