Selina Solutions are useful for students as it helps them in scoring high marks in the examination. The Selina Solutions contain detailed step-by-step explanation of all the problems that come under the Chapter 7, Indices [Exponents], of the Class 9 Selina Textbook.

These solutions are prepared by subject matter experts at BYJU’S, describing the complete method of solving problems. By understanding the concepts used in Selina Solutions for Class 9 Maths, students will be able to clear all their doubts related to “Indices [Exponents]”.

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**Exercise 7(A)**

**1. Evaluate:**

**(i) 3 ^{3} x (243)^{-2/3} x 9^{-1/3}**

**(ii) 5 ^{-4} x (125)^{5/3} ÷ (25)^{-1/2}**

**(iii) (27/125) ^{2/3} x (9/25)^{-3/2}**

**(iv) 7 ^{0} x (25)^{-3/2} – 5^{-3} **

**(v) (16/81) ^{-3/4} x (49/9)^{3/2} ÷ (343/216)^{2/3} **

**Solution:**

(i) 3^{3} x (243)^{-2/3} x 9^{-1/3} = 3^{3} x (3 x 3 x 3 x 3 x 3)^{-2/3} x (3 x 3)^{-1/3}

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

= 3^{3} x (3)^{-10/3} x 3^{-2/3} [As (a^{m})^{n} = a^{mn}]

= 3^{3 – 10/3 – 2/3 }[a^{m} x a^{n} x a^{p} = a^{m + n + p}]

= 3 ^{(9 – 10 – 2)/3}

= 3^{-3/3}

= 3^{-1}

= 1/3

** **

(ii) 5^{-4} x (125)^{5/3} ÷ (25)^{-1/2} = 5^{-4} x (5 x 5 x 5)^{5/3} ÷ (5 x 5)^{-1/2}

= 5^{-4} x (5^{3})^{5/3} ÷ (5^{2})^{-1/2}

= 5^{-4} x (5^{3 x 5/3}) ÷ (5^{2 x -1/2}) [As (a^{m})^{n} = a^{mn}]

= 5^{-4} x 5^{5} ÷ 5^{-1}

= 5^{-4} x 5^{5} x 5^{-(-1)} [As 1/a^{-m} = a^{-(-m)} = a^{m}]

= 5^{(-4 + 5 + 1)} [a^{m} x a^{n} x a^{p} = a^{m + n + p}]

= 5^{2}

= 25

(iii) (27/125)^{2/3} x (9/25)^{-3/2} = (3 x 3 x 3/5 x 5 x 5)^{2/3} x (3 x 3/5 x 5)^{-3/2}

= (3^{3}/5^{3})^{2/3} x (3^{2}/5^{2})^{-3/2}

= (3/5)^{3 x 2/3} x (3/5)^{2 x -3/2} [As (a^{m})^{n} = a^{mn}]

= (3/5)^{2} x (3/5)^{-3}

= (3/5)^{2 – 3} [a^{m} x a^{n} = a^{m + n}]

= (3/5)^{-1}

^{ } = 5/3

(iv) 7^{0} x (25)^{-3/2} – 5^{-3} = 1 x (5 x 5)^{-3/2} – 5^{-3} [As a^{0} = 1]

= (5^{2})^{-3/2} – 5^{-3}

= (5)^{2 x -3/2} – 5^{-3 } [As (a^{m})^{n} = a^{mn}]

** **= 5^{-3} – 5^{-3}

= 1/5^{3} – 1/5^{3}

= 0

(v) (16/81)^{-3/4} x (49/9)^{3/2} ÷ (343/216)^{2/3}

= (2 x 2 x 2 x 2/3 x 3 x 3 x 3)^{-3/4} x (7 x 7/3 x 3)^{3/2} ÷ (7 x 7 x 7/6 x 6 x 6)^{2/3}

= (2^{4}/3^{4})^{-3/4} x (7^{2}/3^{2})^{3/2} ÷ (7^{3}/6^{3})^{2/3}

= (2/3)^{4 x -3/4} x (7/3)^{2 x 3/2} ÷ (7/6)^{3 x 2/3}

= (2/3)^{-3} x (7/3)^{3} ÷ (7/6)^{2 } [As (a^{m})^{n} = a^{mn}]

= [1/(2/3)^{3} x (7/3)^{3}]/ (7/6)^{2} [As a^{-m} = 1/a^{m}]

= [(3/2)^{3} x (7/3)^{3}] x (7/6)^{-2}

= (3/2)^{3} x (7/3)^{3} x (6/7)^{2} [As (a/b)^{-m} = (b/a)^{m}]

= 3/2 x 3/2 x 3/2 x 7/3 x 7/3 x 7/3 x 6/7 x 6/7

= (7 x 3 x 3)/2

= 63/2

= 31.5

**2. Simplify:**

**(i) (8x ^{3} ÷ 125y^{3})^{2/3}**

**(ii) (a + b) ^{-1}. (a^{-1} + b^{-1})**

**(iii) **

**(iv) (3x ^{2})^{-3} **×

**(x**

^{9})^{2/3}**Solution:**

(i) (8x^{3} ÷ 125y^{3})^{2/3} = (8x^{3}/125y^{3})^{2/3}

= (2x × 2x × 2x/5y × 5y × 5y)^{2/3}

= (2x^{3}/5y^{3})^{2/3}

= (2x/5y)^{3 x 2/3}

= (2x/5y)^{2}

= 4x^{2}/25y^{2}

(ii) (a + b)^{-1}. (a^{-1} + b^{-1}) = 1/(a + b) x (1/a + 1/b)

= 1/(a + b) x (b + a)/ab

= 1/ab

** **

(iii)

= (5^{1} x 19)/5

= 19

** **

(iv) (3x^{2})^{-3} × (x^{9})^{2/3} = 3^{-3} × (x^{2})^{-3} × (x)^{9 x 2/3}

= 3^{-3 }× (x)^{2 x -3} × (x)^{9 x 2/3}

= 3^{-3} × x^{-6} × x^{6}

= 3^{-3} x 1

= 1/27

** **

** 3. Evaluate:**

**(i) √¼ + (0.01) ^{-1/2} – (27)^{2/3}**

**(ii) (27/8) ^{2/3} – (1/4)^{-2} + 5^{0}**

** **

**Solution:**

(i) √¼ + (0.01)^{-1/2} – (27)^{2/3} = √(½ x ½) + (0.1 x 0.1)^{-1/2} – (3 x 3 x 3)^{2/3}

= √(½)^{2} + (0.1^{2})^{ -1/2} – (3^{3})^{2/3}

^{ } = ½ + (0.1)^{2 x -1/2} – (3)^{3 x} ^{2/3}

= ½ + (0.1)^{-1} – (3)^{2}

= ½ + 1/0.1 – 3^{2}

= ½ + 1/(1/10) – 9

= ½ + 10 – 9

= ½ + 1

= 3/2

** **

(ii)** **(27/8)^{2/3} – (1/4)^{-2} + 5^{0} = (3 x 3 x 3/2 x 2 x 2)^{2/3} – (1/2 x 1/2)^{-2} + 5^{0}

= (3^{3}/2^{3})^{2/3} – (½)^{2})^{-2} + 1

= (3/2)^{3 x 2/3} – (½)^{-4} + 1

= (3/2)^{2} – (½)^{-4} + 1

= (3/2)^{2} – 2^{4} + 1

= (3 x 3)/(2 x 2) – (2 x 2 x 2 x 2) + 1

= 9/4 – 16 + 1

= (9 – 64 + 4)/4

= -51/4

** **

**4. Simplify each of the following and express with positive index:**

**(i) (3 ^{-4}/2^{-8})^{1/4}**

**(ii) (27 ^{-3}/9^{-3})^{1/5}**

**(iii) (32) ^{-2/5} ÷ (125)^{-2/3}**

**(iv) [1 – {1 – (1 – n) ^{-1}}^{-1}]^{-1}**

**Solution:**

(i) (3^{-4}/2^{-8})^{1/4} = (2^{8}/3^{4})^{1/4}

= (2^{8})^{1/4}/(3^{4})^{1/4}

= (2)^{8/4}/(3)^{4/4}

= 2^{2}/3

= 4/3

(ii) (27^{-3}/9^{-3})^{1/5} = (9^{3}/27^{3})^{1/5}

= [(3 x 3)^{3}/(3 x 3 x 3)^{3}]^{1/5}

= [(3^{2})^{3}/(3^{3})^{3}]^{1/5}

= [(3)^{2 x 3}/(3)^{3 x 3}]^{1/5}

= [(3)^{6}/(3)^{9}]^{1/5}

= [(3)^{6-9}]^{1/5}

= (3)^{-3 x 1/5}

= (3)^{-3/5}

^{ } = 1/3^{3/5}

** **

(iii) (32)^{-2/5} ÷ (125)^{-2/3} = (32)^{-2/5}/(125)^{-2/3}

= (125)^{2/3}/(32)^{2/5}

= (5 x 5 x 5)^{2/3}/(2 x 2 x 2 x 2 x 2)^{2/5}

= (5^{3})^{2/3}/(2^{5})^{2/5}

= 5^{3 x 2/3}/2^{5 x 2/5}

= 5^{2}/2^{2}

= 25/4

(iv) [1 – {1 – (1 – n)^{-1}}^{-1}]^{-1} = [1 – {1 – 1/(1 – n)}^{-1}]^{-1}

= [1 – {((1 – n) – 1)/(1 – n)}^{-1}]^{-1}

= [1 – {– n/(1 – n)}^{-1}]^{-1}

= [1 – {– (1 – n)/n}]^{-1}

= [1 + (1 – n)/n]^{-1}

= [(n + 1 – n)/n]^{-1}

= [1/n]^{-1}

= n

**5. If 2160 = 2 ^{a}. 3^{b}. 5^{c}, find a, b and c. Hence, calculate the value of 3^{a} x 2^{-b} x 5^{-c}.**

**Solution:**

We have,

2160 = 2^{a }x 3^{b }x 5^{c}

(2 x 2 x 2 x 2) x (3 x 3 x 3) x 5 = 2^{a }x 3^{b }x 5^{c}

2^{4 }x 3^{3 }x 5^{1} = 2^{a }x 3^{b }x 5^{c}

⇒ 2^{a }x 3^{b }x 5^{c} = 2^{4 }x 3^{3 }x 5^{1}

Comparing the exponents of 2, 3 and 5 on both sides, we get

a = 4, b = 3 and c = 1

Hence, the value

3^{a} x 2^{-b} x 5^{-c} = 3^{4 }x 2^{-3 }x 5^{-1}

= (3 x 3 x 3 x 3) x (½ x ½ x ½) x 1/5

= 81 x 1/8 x 1/5

= 81/40

**6. If 1960 = 2 ^{a}. 5^{b}. 7^{c}, calculate the value of 2^{-a}. 7^{b}. 5^{-c}.**

**Solution:**

We have,

1960 = 2^{a }x 5^{b }x 7^{c}

(2 x 2 x 2) x 5 x (7 x 7) = 2^{a }x 5^{b }x 7^{c}

2^{3 }x 5^{1 }x 7^{2} = 2^{a }x 5^{b }x 7^{c}

⇒ 2^{a }x 5^{b }x 7^{c} = 2^{2 }x 5^{1 }x 7^{2}

Comparing the exponents of 2, 5 and 7 on both sides, we get

a = 3, b = 1 and c = 2

Hence, the value

2^{-a}. 7^{b}. 5^{-c} = 2^{-3} x 7^{1 }x 5^{-2}

= (½ x ½ x ½) x 7 x (1/5 x 1/5)

= 1/8 x 7 x 1/25

= 7/200

** **

**7. Simplify:**

**(i) **

**(ii) **

**Solution:**

(i)

** **

(ii)

**8. Show that:**

**(a ^{m}/a^{-n})^{m-n} × (a^{n}/a^{-l})^{n-l} × (a^{l}/a^{-m})^{l-m} = 1**

**Solution:**

Taking the L.H.S., we have

(a^{m}/a^{-n})^{m-n} × (a^{n}/a^{-l})^{n-l} × (a^{l}/a^{-m})^{l-m}

= (a^{m} × a^{n})^{m-n} × (a^{n} ×a^{l})^{n-l} × (a^{l} × a^{m})^{l-m}

= (a^{m+n})^{ m-n} × (a^{n+l})^{n-l} × (a^{l+m})^{l-m}

** **

= a^{0}

= 1

**9. If a = x ^{m+n}. x^{l}; b = x^{n+l}.^{ }x^{m} and c = x^{l+m}. x^{n},**

**Prove that: a ^{m-n}. b^{n-l}. c^{l-m} = 1**

**Solution:**

We have,

a = x^{m+n}. x^{l}

b = x^{n+l}.^{ }x^{m}

c = x^{l+m}. x^{n}

Now,

Considering the L.H.S.,

a^{m-n}. b^{n-l}. c^{l-m}

= (x^{m+n}. x^{l})^{m-n}. (x^{n+l}.^{ }x^{m})^{n-l}. (x^{l+m}. x^{n})^{l-m}

= [x^{(m+n)(m-n)}. x^{l(m-n)}]. [x^{(n+l)(n-l)}. x^{m(n-l)}]. [x^{(l+m)(l-m)}. x^{n(l-m)}]

= x^{0}

= 1

= R.H.S

– Hence proved.

**10. Simplify:**

**(i) **

** **

**(ii) **

**Solution:**

(i)

= x^{0}

= 1

** **

**(ii)**

**Exercise 7(B)**

**1. Solve for x:**

**(i) 2 ^{2x+1} = 8**

**(ii) 2 ^{5x-1} = 4 ×_{ }2^{3x + 1}**

**(iii) 3 ^{4x + 1} = (27)^{x + 1}**

**(iv) (49) ^{x + 4} = 7^{2 }× (343)^{x + 1}**

**Solution:**

(i) We have, 2^{2x+1} = 8

⇒2^{2x+1} = 2^{3}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

2x + 1 = 3

2x = 3 – 1

2x = 2

x = 2/2

x = 1

(ii) We have, 2^{5x-1} = 4 ×_{ }2^{3x + 1}

⇒ 2^{5x-1} = 2^{2} ×_{ }2^{3x + 1}

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

2^{5x-1} =_{ }2^{3x + 3}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

5x – 1 = 3x + 3

5x – 3x = 3 + 1

2x = 4

x = 4/2

x = 2

(iii) We have, 3^{4x + 1} = (27)^{x + 1}

3^{4x + 1} = (3^{3})^{x + 1}

3^{4x + 1} = (3)^{3x + 3}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

4x + 1 = 3x + 3

4x – 3x = 3 – 1

x = 2

(iv) We have, (49)^{x + 4} = 7^{2 }× (343)^{x + 1}

(7 x 7)^{x + 4} = 7^{2 }× (7 x 7 x 7)^{x + 1}

(7^{2})^{x + 4} = 7^{2 }× (7^{3})^{x + 1}

(7)^{2x + 8} = (7)^{3x + 3 + 2}

(7)^{2x + 8} = (7)^{3x + 5}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

2x + 8 = 3x + 5

3x – 2x = 8 – 5

x = 3

**2. Find x, if:**

**(i) 4 ^{2x }= 1/32**

**(ii) √(2 ^{x+3}) = 16**

**(iii) [√(⅗)] ^{x+1} = 125/27 **

**(iv) [∛(⅔)] ^{x-1} = 27/8**

**Solution:**

(i) We have, 4^{2x }= 1/32

(2 × 2)^{2x} = 1/(2 × 2 × 2 × 2 × 2)

(2^{2})^{2x} = 1/2^{5}

2^{4x} = 2^{-5}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

4x = -5

x = -5/4

(ii) We have, √(2^{x+3}) = 16

(2^{x+3})^{1/2} = (2 × 2 × 2 × 2)

2^{(x+3)/2} = 2^{4}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

(x + 3)/2 = 4

x + 3 = 8

x = 8 – 3

x = 5

(iii) We have, [√(⅗)]^{x+1} = 125/27

[(⅗)^{1/2}]^{x+1} = (5 × 5 × 5)/(3 × 3 × 3)

(⅗)^{(x+1)/2} = 5^{3}/3^{3}

(⅗)^{(x+1)/2} = (5/3)^{3}

(⅗)^{(x+1)/2} = (3/5)^{-3}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

(x + 1)/2 = -3

x + 1 = -6

x = -6 – 1

x = -7

(iv) We have, [∛(⅔)]^{x-1} = 27/8

[(⅔)^{1/3}]^{x-1} = (3 × 3 × 3)/(2 × 2 × 2)

(⅔)^{(x-1)/3} = (3/2)^{3}

(⅔)^{(x-1)/3} = (2/3)^{-3}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

(x – 1)/3 = -3

x – 1 = -9

x = -9 + 1

x = -8

**3. Solve:**

**(i) 4 ^{x-2} – 2^{x+1} = 0**

**(ii) _{}**

**Solution:**

(i) We have,

4^{x-2} – 2^{x+1} = 0

(2^{2})^{x-2} – 2^{x+1} = 0

2^{2x-4} – 2^{x+1} = 0

2^{2x-4} = 2^{x+1}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

2x – 4 = x + 1

2x – x = 4 + 1

x = 5

(ii) We have,

_{}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

x^{2} = x+ 2

x^{2} – x – 2 = 0

On factorization, we get

x^{2} – 2x + x – 2 = 0

x(x – 2 + 1(x – 2) = 0

(x + 1)(x – 2) = 0

So, either (x + 1) = 0 or (x – 2) = 0

Thus, x = -1 or 2

**4. Solve:**

**(i) 8 × 2 ^{2x} + 4_{ }× 2^{x+1} = 1 + 2^{x}**

**(ii) 2 ^{2x} + 2^{x+2} – 4 × 2^{3} = 0**

**(iii) (√3) ^{x-3}_{ }= (∜3)^{x+1}**

**Solution:**

(i) We have, 8 × 2^{2x} + 4_{ }× 2^{x+1} = 1 + 2^{x}

8 × (2^{x})^{2} + 4_{ }× (2^{x}) × 2^{1} = 1 + 2^{x}

Let us substitute 2^{x} = t

Then,

8 × t^{2} + 4_{ }× t × 2 = 1 + t

8t^{2} + 8t = 1 + t

8t^{2} + 8t – t – 1 = 0

8t^{2} + 7t – 1 = 0

8t^{2} + 8t – t – 1 = 0

8t(t + 1) – 1(t + 1) = 0

(8t – 1)(t + 1) = 0

So, either 8t – 1 = 0 or t + 1 = 0

Thus, t = 1/8 or -1

Now, we have

2^{x} = t

So,

2^{x} = 1/8 or 2^{x} = -1

The equation, 2^{x} = -1 is not possible

Hence, for 2^{x} = 1/8

2^{x} = 1/(2 × 2 × 2)

2^{x} = 1/2^{3}

2^{x} = 2^{-3}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

x = -3

** **

(ii) We have,

2^{2x} + 2^{x+2} – 4 × 2^{3} = 0

2^{2x} + 2^{x+2} – 2^{2} × 2^{3} = 0

(2^{x})^{2} + 2^{x}.2^{2} – 2^{3+2} = 0

(2^{x})^{2} + 2^{x}.2^{2} – 2^{5} = 0

Now, let’s assume 2^{x} = t

So, the above equation becomes

(t)^{2} + t.2^{2} – 2^{5} = 0

t^{2} + 4t – 32 = 0

On factorization, we have

t^{2} + 8t – 4t – 32 = 0

t(t + 8) – 4(t + 8) = 0

(t – 4)(t + 8) = 0

So, either (t – 4) = 0 or (t + 8) = 0

Thus, t = 4 or -8

Now, we have t = 2^{x}

So,

2^{x} = 4 or 2^{x} = -8

The equation, 2^{x} = -8 is not possible

Hence, for

2^{x} = 4

2^{x} = 2^{2}

On comparing the exponents, we get

x = 2

** **

(iii) (√3)^{x-3}_{ }= (∜3)^{x+1}

(3^{1/2})^{x-3}_{ }= (3^{1/4})^{x+1}

3^{(x-3)/2}_{ }= 3^{(x+1)/4}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

(x – 3)/2 = (x + 1)/4

2(x – 3) = (x + 1)

2x – 6 = x + 1

2x – x = 6 + 1

x = 7

**5. Find the values of m and n if:**

**4 ^{2m} = (∛16)^{-6/n} = (√8)^{2}**

**Solution:**

We have, 4^{2m} = (∛16)^{-6/n} = (√8)^{2}

Now, considering

4^{2m} = (√8)^{2}

(2^{2})^{2m} = (8^{1/2})^{2}

2^{4m} = 8^{1/2 x 2}

2^{4m} = 8

⇒ 2^{4m} = 2^{3}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

4m = 3

m = ¾

Now, from the given considering

(∛16)^{-6/n} = (√8)^{2}

(16^{1/3})^{-6/n} = (8^{1/2})^{2}

(16)^{1/3 x -6/n} = 8^{1/2 x 2}

(16)^{-2/n} = 8

(2 x 2 x 2 x 2)^{-2/n} = (2 x 2 x 2)

(2)^{4 x -2/n} = 2^{3}

(2)^{-8/n} = 2^{3}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

-8/n = 3

n = -8/3

Therefore, the value of m and n are ¾ and -8/3

** **

**6. Solve x and y if:**

**(√32) ^{x} ÷ 2^{y+1} = 1 and 8^{y} – 16^{4 – x/2} = 0**

**Solution:**

Consider the equation, (√32)^{x} ÷ 2^{y+1} = 1

(√(2 × 2 × 2 × 2 × 2))^{x} ÷ 2^{y+1} = 1

(√2^{5})^{x} ÷ 2^{y+1} = 1

(2^{5})^{1/2 × x} ÷ 2^{y+1} = 1

2^{5x/2} ÷ 2^{y+1} = 1

(2^{5x/2})/(2^{y+1}) = 1

2^{5x/2 – (y+1)} = 2^{0}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

5x/2 – (y+1) = 0

5x/2 – y – 1 = 0

5x – 2y – 2 = 0 … (i)

Next, let’s consider 8^{y} – 16^{4 – x/2} = 0

(2 × 2 × 2)^{y} – (2 × 2 × 2 × 2)^{4 – x/2} = 0

(2^{3})^{y} – (2^{4})^{4 – x/2} = 0

2^{3y} – 2^{16 – 2x} = 0

2^{3y} = 2^{16 – 2x}

Now, if the bases are equal, then the powers must be equal

So, on comparing the exponents, we get

3y = 16 – 2x

2x + 3y – 16 = 0 … (ii)

On solving equations (i) and (ii),

By manipulating by (i) × 3 + (ii) × 2, we have

15x – 6y – 6 = 0

4x + 6y – 32 = 0

———————–

19x – 38 = 0

x = 38/19

x = 2

Now, substituting the value of x in (i)

5(2) – 2y – 2 = 0

10 – 2y – 2 = 0

8 = 2y

y = 8/2

y = 4_{ }

Therefore, the values of x and y are 2 and 4 respectively

**7. Prove that:**

**(i) (x ^{a}/x^{b})^{a+b-c}. (x^{b}/x^{c})^{b+c-a}. (x^{c}/x^{a})^{c+a-b} = 1**

**(ii) x ^{a(b-c)}/x^{b(a-c)} ÷ (x^{b}/x^{a})^{c} = 1**

**Solution:**

(i) Taking L.H.S, we have

(x^{a}/x^{b})^{a+b-c}. (x^{b}/x^{c})^{b+c-a}. (x^{c}/x^{a})^{c+a-b}

= (x^{a}/x^{b})^{a+b-c}. (x^{b}/x^{c})^{b+c-a}. (x^{c}/x^{a})^{c+a-b}

= x^{(a-b)(a+b-c)}. x^{(b-c)(b+c-a)}. x^{(c-a)(c+a-b)}

_{}

= x^{0}

= 1

= R.H.S

** **

(ii) Taking L.H.S, we have

x^{a(b-c)}/x^{b(a-c)} ÷ (x^{b}/x^{a})^{c}

= x^{a(b-c) – b(a-c)} ÷ x^{c(b-a)}

= x^{a(b-c) – b(a-c)}/x^{c(b-a)}

= x^{a(b-c) – b(a-c) – c(b-a)}

= x^{ab-ac-ba+bc-cb+ac}

= x^{0}

= 1

= R.H.S

**8. If a ^{x} = b, b^{y} = c and c^{z} = a, prove that: xyz = 1.**

**Solution:**

We have, a^{x} = b, b^{y} = c and c^{z} = a

Now, considering

a^{x} = b

On raising to the power yz on both sides, we get

(a^{x})^{yz} = (b)^{yz}

(a)^{xyz} = (b^{y})^{z}

(a)^{xyz} = (c)^{z} [As, b^{y} = c]

a^{xyz} = a

a^{xyz} = a^{1} [As, c^{z} = a]

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

Hence, xyz = 1

**9. If a ^{x} = b^{y} = c^{z} and b^{2} = ac, prove that: y = 2az/(x + z).**

**Solution:**

Let’s assume a^{x} = b^{y} = c^{z} = k

So,

a = k^{1/x}; b = k^{1/y} and c = k^{1/z}

Now,

It’s given that b^{2} = ac

⇒ (k^{1/y})^{2} = (k^{1/x}) × (k^{1/z})

(k^{2/y}) = k^{1/x + 1/z}

k^{2/y} = k^{(z+x)/xz}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

2/y = (z + x)/xz

2xz = y(z + x)

Hence,

y = 2xz/(x + z)

**10. If 5 ^{-p} = 4^{-q} = 20^{r}, show that: 1/p + 1/q + 1/r = 0.**

**Solution:**

Let’s assume 5^{-p} = 4^{-q} = 20^{r} = k

Then, as

5^{-p} = k ⇒ 5 = k^{-1/p}

4^{-q} = k ⇒ 4 = k^{-1/q}

20^{r} = k ⇒ 20 = k^{1/r}

Now, we know

5 x 4 = 20

(k^{-1/p})^{ }x (k^{-1/q}) = k^{1/r}

k^{-1/p – 1/q} = k^{1/r}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

-1/p – 1/q = 1/r

Hence,

1/p + 1/q + 1/r = 0** **

** **

**11. If m ≠ n and (m + n) ^{-1} (m^{-1} + n^{-1}) = m^{x}n^{y}, show that: x + y + 2 = 0**

**Solution:**

Given equation,

(m + n)^{-1} (m^{-1} + n^{-1}) = m^{x}n^{y}

1/(m + n) × (1/m + 1/n) = m^{x}n^{y}

1/(m + n) × (m + n)/mn = m^{x}n^{y}

1/mn = m^{x}n^{y}

m^{-1}n^{-1} = m^{x}n^{y}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

x = -1 and y = -1

Substituting the values of x and y in the equation x + y + 2 = 0, we have

(-1) + (-1) + 2 = 0

0 = 0

L.H.S = R.H.S

Therefore, x + y + 2 = 0** **

** **

**12. If 5 ^{x + 1} = 25^{x – 2}, find the value of 3^{x – 3} × 2^{3 – x}**

**Solution:**

We have, 5^{x + 1} = 25^{x – 2}

5^{x + 1} = (5^{2})^{x – 2}

5^{x + 1} = 5^{2x – 4}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

x + 1 = 2x – 4

2x – x = 4 + 1

x = 5

Hence, the value of

3^{x – 3} × 2^{3 – x} = 3^{5 – 3} × 2^{3 – 5}

= 3^{2} × 2^{-2}

= 9 × ¼

= 9/4

**13. If 4 ^{x + 3} = 112 + 8 × 4^{x}, find the value of (18x)^{3x}.**

**Solution:**

We have,

4^{x + 3} = 112 + 8 × 4^{x}

4^{x }.4^{3} = 112 + 8 × 4^{x}

Let’s assume 4^{x} = t

Then,

t^{ }.4^{3} = 112 + 8 × t

64t = 112 + 8t

64t – 8t = 112

56t = 112

t = 112/56

t = 2

But we have taken 4^{x} = t

So, 4^{x} = 2

(2^{2})^{x} = 2^{1}

2^{2x} = 2^{1}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

2x = 1

x = ½

Now, the value of (18x)^{3x} will be

= (18 × ½)^{3 × 1/2 }

= (9)^{3/2}

= (3^{2})^{3/2}

= 3^{3}

= 27

**14. Solve for x:**

**(i) 4 ^{x-1} × (0.5)^{3 – 2x} = (1/8)^{-x}**

**Solution:**

We have,

4^{x-1} × (0.5)^{3 – 2x} = (1/8)^{-x}

(2^{2})^{x-1} × (1/2)^{3 – 2x} = (1/2^{3})^{-x}

(2)^{2x-1} × (2)^{-(3 – 2x)} = (2^{-3})^{-x}

(2)^{2x-2} × (2)^{2x-3} = (2)^{3x}

2^{(2x-2) + (2x-3)} = (2)^{3x}

2^{4x-5} = 2^{3x}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

4x – 5 = 3x

4x – 3x = 5

x = 5

** **

**(ii) (a ^{3x + 5})^{2} **×

**(a**

^{x})^{4}= a^{8x + 12}**Solution:**

We have,

(a^{3x + 5})^{2 }× (a^{x})^{4} = a^{8x + 12}

a^{6x + 10} × a^{4x} = a^{8x + 12}

a^{6x + 10 + 4x} = a^{8x + 12}

a^{10x + 10} = a^{8x + 12}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

10x + 10 = 8x + 12

10x – 8x = 12 – 10

2x = 2

x = 1** **

**(iii) (81) ^{3/4} – (1/32)^{-2/5} + x(1/2)^{-1} × 2^{0} = 27**

**Solution:**

We have,

(81)^{3/4} – (1/32)^{-2/5} + x(1/2)^{-1} × 2^{0} = 27

(3^{4})^{3/4} – (1/2^{5})^{-2/5} + x(1/2)^{-1} × 2^{0} = 27

(3^{4})^{3/4} – (2^{-5})^{-2/5} + x(2^{-1})^{-1} × 2^{0} = 27

3^{3} – 2^{2} + 2x × 1 = 27

27 – 4 + 2x = 27

2x + 23 = 27

2x = 27 – 23

2x = 4

x = 4/2

x = 2** **

** **

**(iv) 2 ^{3x + 3} = 2^{3x + 1} + 48**

**Solution:**

We have,

2^{3x + 3} = 2^{3x + 1} + 48

2^{3x + 3} – 2^{3x + 1} = 48

2^{3x}(2^{3} – 2^{1}) = 48

2^{3x}(8 – 2) = 48

2^{3x }× 6 = 48

2^{3x} = 48/6

2^{3x} = 8

2^{3x} = 2^{3}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

3x = 3

x = 1

**(v) 3(2 ^{x} + 1) – 2^{x+2} + 5 = 0**

**Solution:**

We have,

3(2^{x} + 1) – 2^{x+2} + 5 = 0

3 × 2^{x} + 3 – 2^{x}.2^{2} + 5 = 0

2^{x}(3 – 2^{2}) + 5 + 3 = 0

2^{x}(3 – 4) + 8 = 0

-2^{x }+ 8 = 0

2^{x} = 8

2^{x} = 2^{3}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

x = 3

** **

**(vi) 9 ^{x+2 }= 720 + 9^{x}**

**Solution:**

We have,

9^{x+2 }= 720 + 9^{x}

9^{x+2} – 9^{x} = 720

9^{x} (9^{2} – 1) = 720

9^{x} (81 – 1) = 720

9^{x} (80) = 720

9^{x} = 9

9^{x} = 9^{1}

Therefore, x = 1

**Exercise 7(C)**

**1. Evaluate: **

**(i) 9 ^{5/2} – 3 x 8^{0} – (1/81)^{-1/2}**

**(ii) (64) ^{2/3} – ∛125 – 1/2^{-5} + (27)^{-2/3} x (25/9)^{-1/2}**

**(iii) [(-2/3) ^{-2}]^{3} x (1/3)^{-4} x 3^{-1} x 1/6**

**Solution:**

(i) We have, 9^{5/2} – 3 x 8^{0} – (1/81)^{-1/2}

= (3^{2})^{5/2} – 3 x 1 – (1/9^{2})^{-1/2}

= 3^{2 x 5/2} – 3 – (1/9)^{-2 x ½}

= 3^{5} – 3 – (1/9)^{-1}

= (3 x 3 x 3 x 3 x 3) – 3 – (9^{-1})^{-1}

= 243 – 3 – 9

= 231

(ii) We have, (64)^{2/3} – ∛125 – 1/2^{-5} + (27)^{-2/3} x (25/9)^{-1/2}

= (4 x 4 x 4)^{2/3} – (5 x 5 x 5)^{1/3} – 1/2^{-5} + (3 x 3 x 3)^{-2/3} x (5 x 5/3 x 3)^{-1/2}

= (4^{3})^{2/3} – (5^{3})^{1/3} – (2^{-1})^{-5} + (3^{3})^{-2/3} x (5^{2}/3^{2})^{-1/2}

= (4)^{3 x 2/3} – (5)^{3 x 1/3} – (2)^{-1 x -5} + (3)^{3 x -2/3} x (5/3)^{2 x -1/2}

= (4)^{2} – 5 – 2^{5} + 3^{-2} x (5/3)^{-1}

= 16 – 5 – 32 + 1/9 x 3/5

= -21 + 1/15

= (-315 + 1)/15

= -314/15

(iii) We have, [(-2/3)^{-2}]^{3} x (1/3)^{-4} x 3^{-1} x 1/6

= (-2/3)^{-6} x (3^{-1})^{-4} x 3^{-1} x ½ x 1/3

= (-3/2)^{6} x 3^{4} x 3^{-1} x 2^{-1} x 3^{-1}

= (-1)^{6} x (3^{6} x 3^{4} x 3^{-1} x 3^{-1}) x (2^{-6} x 2^{-1})

= 1 x 3^{6+4-1-1} x 2^{-6-1}

= 3^{8} x 2^{-7}

= 3^{8}/2^{7}

**2. Simplify:**

**Solution:**

We have,

** **

= (27 – 9)/(81 – 9)

= 18/72

= 1/4

**3. Solve: 3 ^{x-1 }× 5^{2y-3} = 225.**

**Solution:**

Given, 3^{x-1}× 5^{2y-3} = 225

3^{x-1 }× 5^{2y-3} = 9 x 25

3^{x-1 }× 5^{2y-3} = 3^{2} x 5^{2}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

x – 1 = 2 and 2y – 3 = 2

x = 2 + 1 and 2y = 2 + 3

x = 3 and 2y = 5

x = 3 and y = 5/2 = 2.5

**4. If [(a ^{-1}b^{2})/(a^{2}b^{-4})]^{7} ÷ [(a^{3}b^{-5})/(a^{-2}b^{3})] = a^{x}.b^{y}, find x + y.**

**Solution:**

We have,

** **

(b^{42}/a^{21})^{ }÷ (b^{40}/a^{25}) = a^{x}. b^{y}

(b^{42}/a^{21})^{ }× (a^{25}/b^{40}) = a^{x}. b^{y}

b^{42 – 40} × a^{25 – 21} = a^{x}. b^{y}

a^{4} × b^{2} = a^{x}. b^{y}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

x = 4 and y = 2

x + y = 4 + 2 = 6

**5. If 3 ^{x+1} = 9^{x-3}, find the value of 2^{1+x}.**

**Solution:**

We have,

3^{x+1} = 9^{x-3}

3^{x} × 3^{1} = (3^{2})^{x – 3}

3^{x} × 3^{1} = (3)^{2x – 6}

3^{x} = (3)^{2x – 6}/3

3^{x} = (3)^{2x – 6} × 3^{-1}

3^{x} = (3)^{2x – 6 – 1}

3^{x} = 3^{2x – 7}

Now, if the bases are equal, then the powers must be equal

On comparing the exponents, we get

x = 2x – 7

x = 7

Now,

2^{1+x} = 2^{1+7} = 2^{8} = 256

**6. If 2 ^{x} = 4^{y} = 8^{z} and 1/2x + 1/4y + 1/8z = 4, find the value of x.**

**Solution:**

Given, 2^{x} = 4^{y} = 8^{z}

2^{x} = (2^{2})^{y} = (2^{3})^{z}

2^{x} = 2^{2y} = 2^{3z}

On comparing the powers, we get

x = 2y = 3z

y = x/2 and z = x/3

Now,

1/2x + 1/4y + 1/8z = 4

1/2x + 1/4(x/2) + 1/8(x/3) = 4

1/2x + 2/4x + 3/8x = 4

1/2x + 1/2x + 3/8x = 4

(4 + 4 + 3)/8x = 4

11/8x = 4

4 × 8x = 11

32x = 11

x = 11/32

**7. If **

**Show that: m – n = 1**

**Solution:**

We have,

** **

On comparing the powers, we get

m – n = 1

**8. Solve for x: (13) ^{√x} = 4^{4} – 3^{4} – 6.**

**Solution:**

We have,

(13)^{√x} = 4^{4} – 3^{4} – 6

= 256 – 81 – 6

= 169

= 13^{2}

13^{√x} = 13^{2}

On comparing the powers, we get

√x = 2

Squaring on both sides,

x = 2^{2}

x = 4

**9. If 3 ^{4x} = (81)^{-1} and (10)^{1/y} = 0.0001, find the value of 2^{-x }×^{ }16^{y}.**

**Solution:**

We have, 3^{4x} = (81)^{-1} and (10)^{1/y} = 0.0001

3^{4x} = (3^{4})^{-1} and (10)^{1/y} = 1/10000

3^{4x} = 3^{-4} and 10^{1/y} = 1/10^{-4}

3^{4x} = 3^{-4} and 10^{1/y} = 1/10^{4}

3^{4x} = 3^{-4} and 10^{1/y} = 10^{-4}

On comparing the powers, we get

4x = -4 and 1/y = -4

x = -1 and y = -1/4

Now, value of

2^{-x }×^{ }16^{y} = 2^{-(-1) }×^{ }16^{-1/4}

= 2^{1} ×^{ }(2^{4})^{-1/4}

= 2 x 2^{-1}

= 2/2

= 1

** **

**10. Solve: 3(2 ^{x} + 1) – 2^{x+2} + 5 = 0.**

**Solution:**

We have, 3(2^{x} + 1) – 2^{x+2} + 5 = 0

(3 × 2^{x} + 3) – (2^{x} × 2^{2}) + 5 = 0

2^{x}(3 – 2^{2}) + 3 + 5 = 0

2^{x}(3 – 4) + 8 = 0

2^{x}(-1) + 8 = 0

-2^{x} + 8 = 0

2^{x} = 8

2^{x} = 2^{3}

On comparing the powers, we get

x = 3

**11. If (a ^{m})^{n} = a^{m} .a^{n}, find the value of: m(n – 1) – (n – 1)**

**Solution:**

We have, (a^{m})^{n} = a^{m} .a^{n}

a^{mn} = a^{m+n}

On comparing the powers, we get

mn = m + n … (i)

Now,

m(n – 1) – (n – 1) = mn – m – n + 1

= (m + n) – m – n + 1 … [Form (i)]

= 1

**12. If m = ∛15 and n = ∛14, find the value of m – n – 1/(m ^{2} + mn + n^{2})**

**Solution:**

We have,

m = ∛15 and n = ∛14

⇒ m = 15^{3} and n = 14^{3}

** **

**13. Evaluate: **

**Solution:**

We have,

= 6/225 x 25/1

= 6/9

= 2/3

**14. Evaluate:**

**(x ^{q}/x^{r})^{1/qr} × (x^{r}/x^{p})^{1/rp} × (x^{p}/x^{q})^{1/pq}**

**Solution:**

We have,

**15. (i) Prove that: a ^{-1}/(a^{-1} + b^{-1}) + a^{-1}/(a^{-1} – b^{-1}) = 2b^{2}/(b^{2} – a^{2})**

**Solution:**

We have,

= 1/a x ab/(b + a) + 1/a x ab/(b – a)

= b/(b + a) + b/(b – a)

= (b^{2} – ab + b^{2} + ab)/(b^{2} – a^{2})

= 2b^{2}/(b^{2} – a^{2})

= R.H.S

**(ii) Prove that: (a + b + c)/(a ^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}) = abc**

**Solution:**

Taking L.H.S., we have

** **= abc

= R.H.S

**16. Evaluate: 4/(216) ^{-2/3} + 1/(256)^{-3/4 }+ 2/(243)^{-1/5}**

**Solution:**

We have,

= 4/6^{-2} + 1/4^{-3} + 2/3^{-1}

= 4 x 6^{2} + 4^{3} + 2 x 3^{1}

= 4 x 36 + 64 + 2 x 3

= 144 + 64 + 6

= 214

## Selina Solutions for Class 9 Maths Chapter 7- Indices [Exponents]

The Chapter 7, Indices [Exponents], contains 3 exercises and the Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

7.1 Introduction

7.2 Laws of Indices [Exponents]

7.3 Handling positive, fractional, negative and zero indices

7.4 Simplification of expressions

7.5 Solving exponential equations

## Selina Solutions for Class 9 Maths Chapter 7- Indices [Exponents]

The Chapter 7 of class 9 takes the students to a new topic Indices[exponents]. If m is a positive integer, then axaxaxa—- upto m terms, is written as a^{m}; where ‘a’ is called the base and m is called the power (or exponent or index). Read and learn the Chapter 7 of Selina textbook to familiarize with the concepts related to exponents or indices. Learn the Selina Solutions for Class 9 effectively to score high in the examination.