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

Last Updated : 06 Apr, 2021

### 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 × am = an + 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 ÷ am = an – 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 × am = an + 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 × am = an + 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 ÷ am = an – 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 × am = an + 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 × am = an + 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 52x ÷ 5-3 = 55

Solution:

52x ÷ 5-3 = 55

52x + 3 = 55 (an ÷ am = an – m)

When the bases are same, exponents are equated

2x + 3 = 5

Transposing 3

2x = 5 – 3

2x = 2

Transposing 2

x = 1

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