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

^{-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/a^{n}= 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}(a^{n}× 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 a^{n}÷ 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}(a^{n}× 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}(a^{n}× 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}(a^{n}÷ 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}(a^{n}× 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}.

^{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/a^{n}= a^{-n})= (3/2)

^{2 + 4}(a^{n}× 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}.

^{-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/a^{n}= 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} = 5^{5}

^{2x}÷ 5

^{-3}= 5

^{5}

**Solution:**

5

^{2x}÷ 5^{-3}= 5^{5}5

^{2x + 3}= 5^{5}(a^{n}÷ 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