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

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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) 3³ 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) 7⁰ x (25)^-3/2 – 5^-3 

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

Solution:

(i) 3³ x (243)^-2/3 x 9^-1/3 = 3³ x (3 x 3 x 3 x 3 x 3)^-2/3 x (3 x 3)^-1/3

= 3³ x (3⁵)^-2/3 x (3²)^-1/3

= 3³ x (3)^-10/3 x 3^-2/3 [As (a^m)ⁿ = a^mn]

= 3^{3 – 10/3 – 2/3} [a^m x aⁿ 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 (5³)^5/3 ÷ (5²)^-1/2

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

= 5^-4 x 5⁵ ÷ 5^-1

= 5^-4 x 5⁵ x 5^-(-1) [As 1/a^-m = a^-(-m) = a^m]

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

= 5²

= 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

= (3³/5³)^2/3 x (3²/5²)^-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 aⁿ = a^{m + n}]

= (3/5)^-1

= 5/3

(iv) 7⁰ x (25)^-3/2 – 5^-3 = 1 x (5 x 5)^-3/2 – 5^-3  [As a⁰ = 1]

= (5²)^-3/2 – 5^-3 

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

= 5^-3 – 5^-3

= 1/5³ – 1/5³

= 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

= (2⁴/3⁴)^-3/4 x (7²/3²)^3/2 ÷ (7³/6³)^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) (8x³ ÷ 125y³)^2/3

(ii) (a + b)^-1. (a^-1 + b^-1)

(iii) 

(iv) (3x²)^-3 × (x⁹)^2/3

Solution:

(i) (8x³ ÷ 125y³)^2/3 = (8x³/125y³)^2/3

= (2x × 2x × 2x/5y × 5y × 5y)^2/3

= (2x³/5y³)^2/3

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

= (2x/5y)²

= 4x²/25y²

(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)

= (5¹ x 19)/5

= 19

(iv) (3x²)^-3 × (x⁹)^2/3 = 3^-3 × (x²)^-3 × (x)^{9 x 2/3}

= 3^-3 × (x)^{2 x -3} × (x)^{9 x 2/3}

= 3^-3 × x^-6 × x⁶

= 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 + 5⁰

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 + (0.1²) ^-1/2 – (3³)^2/3

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

= ½ + (0.1)^-1 – (3)²

= ½ + 1/0.1 – 3²

= ½ + 1/(1/10) – 9

= ½ + 10 – 9

= ½ + 1

= 3/2

(ii) (27/8)^2/3 – (1/4)^-2 + 5⁰ = (3 x 3 x 3/2 x 2 x 2)^2/3 – (1/2 x 1/2)^-2 + 5⁰

= (3³/2³)^2/3 – (½)²)^-2 + 1

= (3/2)^{3 x 2/3} – (½)^-4 + 1

= (3/2)² – (½)^-4 + 1

= (3/2)² – 2⁴ + 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 = (2⁸/3⁴)^1/4

= (2⁸)^1/4/(3⁴)^1/4

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

= 2²/3

= 4/3

(ii) (27^-3/9^-3)^1/5 = (9³/27³)^1/5

= [(3 x 3)³/(3 x 3 x 3)³]^1/5

= [(3²)³/(3³)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

= (5³)^2/3/(2⁵)^2/5

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

= 5²/2²

= 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

2⁴ x 3³ x 5¹ = 2^a x 3^b x 5^c

⇒ 2^a x 3^b x 5^c = 2⁴ x 3³ x 5¹

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 = 3⁴ 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

2³ x 5¹ x 7² = 2^a x 5^b x 7^c

⇒ 2^a x 5^b x 7^c = 2² x 5¹ x 7²

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 7¹ 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 × (aⁿ/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 × (aⁿ/a^-l)^n-l × (a^l/a^-m)^l-m

= (a^m × aⁿ)^m-n × (aⁿ ×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

= a⁰

= 1

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

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. xⁿ

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. xⁿ)^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)]

= x⁰

= 1

= R.H.S

– Hence proved.

10. Simplify:

(i)

(ii) 

Solution:

(i)

= x⁰

= 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} = 7² × (343)^{x + 1}

Solution:

(i) We have, 2^2x+1 = 8

⇒2^2x+1 = 2³

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 = 2² × 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} = (3³)^{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} = 7² × (343)^{x + 1}

(7 x 7)^{x + 4} = 7² × (7 x 7 x 7)^{x + 1}

(7²)^{x + 4} = 7² × (7³)^{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)

(2²)^2x = 1/2⁵

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 = 2⁴

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 = 5³/3³

(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

(2²)^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

x² = x+ 2

x² – x – 2 = 0

On factorization, we get

x² – 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 × 2³ = 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) × 2¹ = 1 + 2^x

Let us substitute 2^x = t

Then,

8 × t² + 4 × t × 2 = 1 + t

8t² + 8t = 1 + t

8t² + 8t – t – 1 = 0

8t² + 7t – 1 = 0

8t² + 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 = 1/2³

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 × 2³ = 0

2^2x + 2^x+2 – 2² × 2³ = 0

(2^x)² + 2^x.2² – 2³⁺² = 0

(2^x)² + 2^x.2² – 2⁵ = 0

Now, let’s assume 2^x = t

So, the above equation becomes

(t)² + t.2² – 2⁵ = 0

t² + 4t – 32 = 0

On factorization, we have

t² + 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 = 2²

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

(2²)^2m = (8^1/2)²

2^4m = 8^{1/2 x 2}

2^4m = 8

⇒ 2^4m = 2³

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} = 2³

(2)^-8/n = 2³

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

(√2⁵)^x ÷ 2^y+1 = 1

(2⁵)^{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)} = 2⁰

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

(2³)^y – (2⁴)^{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)

= x⁰

= 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}

= x⁰

= 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 = a¹ [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 b² = 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 b² = 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^xn^y, show that: x + y + 2 = 0

Solution:

Given equation,

(m + n)^-1 (m^-1 + n^-1) = m^xn^y

1/(m + n) × (1/m + 1/n) = m^xn^y

1/(m + n) × (m + n)/mn = m^xn^y

1/mn = m^xn^y

m^-1n^-1 = m^xn^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} = (5²)^{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}

= 3² × 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 .4³ = 112 + 8 × 4^x

Let’s assume 4^x = t

Then,

t .4³ = 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

(2²)^x = 2¹

2^2x = 2¹

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

= (3²)^3/2

= 3³

= 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

(2²)^x-1 × (1/2)^{3 – 2x} = (1/2³)^-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 × 2⁰ = 27

Solution:

We have,

(81)^3/4 – (1/32)^-2/5 + x(1/2)^-1 × 2⁰ = 27

(3⁴)^3/4 – (1/2⁵)^-2/5 + x(1/2)^-1 × 2⁰ = 27

(3⁴)^3/4 – (2^-5)^-2/5 + x(2^-1)^-1 × 2⁰ = 27

3³ – 2² + 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(2³ – 2¹) = 48

2^3x(8 – 2) = 48

2^3x × 6 = 48

2^3x = 48/6

2^3x = 8

2^3x = 2³

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.2² + 5 = 0

2^x(3 – 2²) + 5 + 3 = 0

2^x(3 – 4) + 8 = 0

-2^x + 8 = 0

2^x = 8

2^x = 2³

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 (9² – 1) = 720

9^x (81 – 1) = 720

9^x (80) = 720

9^x = 9

9^x = 9¹

Therefore, x = 1

Exercise 7(C)

1. Evaluate:

(i) 9^5/2 – 3 x 8⁰ – (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 8⁰ – (1/81)^-1/2

= (3²)^5/2 – 3 x 1 – (1/9²)^-1/2

= 3^{2 x 5/2} – 3 – (1/9)^{-2 x ½}

= 3⁵ – 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

= (4³)^2/3 – (5³)^1/3 – (2^-1)^-5 + (3³)^-2/3 x (5²/3²)^-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 – 2⁵ + 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 3⁴ x 3^-1 x 2^-1 x 3^-1

= (-1)⁶ x (3⁶ x 3⁴ x 3^-1 x 3^-1) x (2^-6 x 2^-1)

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

= 3⁸ x 2^-7

= 3⁸/2⁷

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 = 3² x 5²

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^-1b²)/(a²b^-4)]⁷ ÷ [(a³b^-5)/(a^-2b³)] = a^x.b^y, find x + y.

Solution:

We have,

(b⁴²/a²¹) ÷ (b⁴⁰/a²⁵) = a^x. b^y

(b⁴²/a²¹) × (a²⁵/b⁴⁰) = a^x. b^y

b^{42 – 40} × a^{25 – 21} = a^x. b^y

a⁴ × b² = 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 × 3¹ = (3²)^{x – 3}

3^x × 3¹ = (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 = 2¹⁺⁷ = 2⁸ = 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 = (2²)^y = (2³)^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 = 4⁴ – 3⁴ – 6.

Solution:

We have,

(13)^√x = 4⁴ – 3⁴ – 6

= 256 – 81 – 6

= 169

= 13²

13^√x = 13²

On comparing the powers, we get

√x = 2

Squaring on both sides,

x = 2²

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 = (3⁴)^-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 = 1/10⁴

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

= 2¹ × (2⁴)^-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 × 2²) + 5 = 0

2^x(3 – 2²) + 3 + 5 = 0

2^x(3 – 4) + 8 = 0

2^x(-1) + 8 = 0

-2^x + 8 = 0

2^x = 8

2^x = 2³

On comparing the powers, we get

x = 3

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

Solution:

We have, (a^m)ⁿ = a^m .aⁿ

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/(m² + mn + n²)

Solution:

We have,

m = ∛15 and n = ∛14

⇒ m = 15³ and n = 14³

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) = 2b²/(b² – a²)

Solution:

We have,

= 1/a x ab/(b + a) + 1/a x ab/(b – a)

= b/(b + a) + b/(b – a)

= (b² – ab + b² + ab)/(b² – a²)

= 2b²/(b² – a²)

= R.H.S

(ii) Prove that: (a + b + c)/(a^-1b^-1 + b^-1c^-1 + c^-1a^-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 6² + 4³ + 2 x 3¹

= 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.