Class 8 RD Sharma Solutions – Chapter 2 Powers – Exercise 2.2 | Set 2

Chapter 2 Powers – Exercise 2.2 | Set 1

Question 11. By what number should (5/3)^-2 be multiplied so that the product may be (7/3)^-1?

Solution:

Let the number be x

 (5/3)^-2 × x = (7/3)^-1

1/(5/3)² × x = 1/(7/3) (1/aⁿ = a^-n)

x = (3/7) / (3/5)²

= (3/7) / (9/25)

= (3/7) × (25/9) 

3 is the common factor

= (1/7) × (25/3)

= 25/21

Question 12. Find x, if

(i) (1/4)^-4 × (1/4)^-8 = (1/4)^-4x

Solution:

(1/4)^-4 × (1/4)^-8 = (1/4)^-4x

(1/4)^{-4 – 8} = (1/4)^-4x (aⁿ × a^m = a^{n + m})

(1/4)^-12 = (1/4)^-4x

When the bases are same, exponents are equated

-12 = -4x

x = -12/-4

Transposing -4

= 3

(ii) (-1/2)^-19 ÷ (-1/2)⁸ = (-1/2)^-2x+1

Solution:

(-1/2)^-19 ÷ (-1/2)⁸ = (-1/2)^{-2x + 1}

(1/2)^-19-8 = (1/2)^-2x+1 (we know that aⁿ ÷ a^m = a^{n – m})

(1/2)^-27 = (1/2)^{-2x + 1}

When the bases are same, exponents are equated

-27 = -2x + 1

Transposing 1

-2x = -27 – 1

-2x = -28

Transposing -2

x= -28/-2

= 14

(iii) (3/2)^-3 × (3/2)⁵ = (3/2)^{2x + 1}

Solution:

(3/2)^-3 × (3/2)⁵ = (3/2)^{2x + 1}

(3/2)^-3+5 = (3/2)^{2x + 1} (aⁿ × a^m = a^{n + m})

(3/2)² = (3/2)^{2x + 1}

When the bases are same, exponents are equated

2 = 2x + 1

Transposing 1

2x = 2 – 1

2x = 1

Transposing 2

x = 1/2

(iv) (2/5)^-3 × (2/5)¹⁵ = (2/5)^2+3x

Solution:

(2/5)^-3 × (2/5)¹⁵ = (2/5)^{2 + 3x}

(2/5)^-3+15 = (2/5)^{2 + 3x} (aⁿ × a^m = a^{n + m})

(2/5)¹² = (2/5)^2+3x

When the bases are same, exponents are equated

12 = 2 + 3x

Transposing 2

3x = 12 – 2

3x = 10

Transposing 3

x = 10/3

(v) (5/4)^-x ÷ (5/4)^-4 = (5/4)⁵

Solution:

(5/4)^-x ÷ (5/4)^-4 = (5/4)⁵

(5/4)^-x+4 = (5/4)⁵ (aⁿ ÷ a^m = a^{n – m})

When the bases are same, exponents are equated

-x + 4 = 5

Transposing 4

-x = 5 – 4

-x = 1

x = -1

(vi) (8/3)^2x+1 × (8/3)⁵ = (8/3) ^x+2

Solution:

(8/3)^2x+1 × (8/3)⁵ = (8/3)^x+2

(8/3)^2x+1+5 = (8/3) ^{x + 2} (aⁿ × a^m = a^{n + m})

(8/3)^2x+6 = (8/3) ^x+2

When the bases are same, exponents are equated

2x + 6 = x + 2

Transposing 6 and x

2x – x = -6 + 2

x = -4

Question 13. (i) If x= (3/2)² × (2/3)^-4, find the value of x^-2.

Solution:

x = (3/2)² × (2/3)^-4

= (3/2)² × (3/2)⁴ (1/aⁿ = a^-n)

= (3/2)^{2 + 4} (aⁿ × a^m = a^{n + m})

= (3/2)⁶

x^-2 = ((3/2)⁶)^-2

= (3/2)^-12

= (2/3)¹²

(ii) If x = (4/5)^-2 ÷ (1/4)², find the value of x^-1.

Solution:

x = (4/5)^-2 ÷ (1/4)²

= (5/4)² ÷ (1/4)² (1/aⁿ = a^-n)

= (5/4)² × (4/1)²

= 25/16 × 16

16 is the common factor

= 25

x^-1 = 1/25

Question 14. Find the value of x for which 5^2x ÷ 5^-3 = 5⁵

Solution:

5^2x ÷ 5^-3 = 5⁵

5^{2x + 3} = 5⁵ (aⁿ ÷ a^m = a^{n – m})

When the bases are same, exponents are equated

2x + 3 = 5

Transposing 3

2x = 5 – 3

2x = 2

Transposing 2

x = 1

My Personal Notes arrow_drop_up

Save