Frank Solutions for Class 9 Maths Chapter 9 Indices

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1. Evaluate the following:

(i) 6⁰

(ii) (1 / 2)^-3

(iii)

(iv)

(v) (0.008)^{(2 / 3)}

(vi) (0.00243)^{(-3 / 5)}

(vii)

(viii)

Solution:

(i) 6⁰ = 1

(ii) (1 / 2)^-3 = (2)³

We get,

(1 / 2)^-3 = 8

(iii)
= 2⁶

We get,

= 64

(iv)
= 3⁶

We get,

= 729

(v) (0.008)^{(2 / 3)} = (0.2³)^{(2 / 3)}

= (0.2)^{3 × 2 / 3}

= (0.2)²

We get,

(0.008)^{(2 / 3)} = 0.04

(vi) (0.00243)^{(-3 / 5)} = {1 / (0.00243)^{(3 / 5)}}

= {1 / (0.3⁵)^{(3 / 5)}}

= {1 / (0.3)³}

We get,

(0.00243)^{(-3 / 5)} = (1 / 0.027)

(vii)
=

=

= 5^{6 × 1 / 6}

We get,

= 5

(viii)
= (64 / 27)^{(2 / 3)}

= (4 / 3)^{3 × 2 / 3}

= (4 / 3)²

We get,

= (16 / 9)

2. Evaluate the following:

(a) 9⁴ ÷ 27^{(-2 / 3)}

(b) 7^-4 × (343)^{(2 / 3)} ÷ (49)^{(- 1 / 2)}

(c) (64 / 216)^{(2 / 3)} × (16 / 36)^{(- 3 / 2)}

Solution:

(a) 9⁴ ÷ 27^{(-2 / 3)} = {(3)²}⁴ ÷ {(3)³}^{(-2 / 3)}

On further calculation, we get,

= (3)^{2 × 4} ÷ (3)^{3 × (- 2 / 3)} [Using (a^m)ⁿ = a^mn]

= (3)⁸ ÷ (3)^-2

= (3)^{8 – (-2)} [Using a^m ÷ aⁿ = a^{m – n}]

= (3)^{8 + 2}

= (3)¹⁰

This can be written as,

= (3)^{2 × 5}

= {(3)²}⁵

= (9)⁵

We get,

= 59049

(b) 7^-4 × (343)^{(2 / 3)} ÷ (49)^{(- 1 / 2)}

On calculating further, we get,

= 7^-4 × (7³)^{(2 / 3)} ÷ (7²)^{(- 1 / 2)}

= 7^-4 × 7^{3 × 2 / 3} ÷ 7^{2 × (- 1 / 2)}

= 7^-4 × 7² ÷ 7^-1

= 7^{-4 + 2 – (-1)} [Using a^m × aⁿ = a^{m + n} and a^m ÷ aⁿ = a^{m – n}]

= 7^{-4 + 2 + 1}

We get,

= 7^-1

= (1 / 7) [Using a^-m = (1 / a^m)]

(c) (64 / 216)^{(2 / 3)} × (16 / 36)^{(- 3 / 2)}

On further calculation, we get,

= {(2⁶)^{2 / 3} / (6³)^{2 / 3}} × {(2⁴)^{-3 / 2} / (6²)^{– 3 / 2}}

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

= {(2)^{2 × 2} / (6)²} × {(2)^{2 × (- 3)} / (6)^{– 3}}

= {(2)⁴ / (6)²} × {(2)^-6 / (6)^-3}

We get,

= {(2)⁴ / (6)²} × {(6)³ / (2)⁶}

= {(2)⁴ / (2)⁶} × {(6)³ / (6)²}

= (2)^{4 – 6} × (6)^{3 – 2}

= (2)^-2 × (6)¹

= (1 / 2²) × 6

= (1 / 4) × 6

We get,

= (3 / 2)

3. Write each of the following in the simplest form:

(a) (a³)⁵ × a⁴

(b) a² × a³ ÷ a⁴

(c) a^{1 / 3} ÷ a^{-2 / 3}

(d) a^-3 × a² × a⁰

(e) (b^-2 – a^-2) ÷ (b^-1 – a^-1)

Solution:

(a) (a³)⁵ × a⁴ = (a)^{3 × 5} × a⁴

= (a)¹⁵ × a⁴

= a^{15 + 4}

We get,

= a¹⁹

(b) a² × a³ ÷ a⁴ = a^{2 + 3 – 4}

We get,

= a¹

= a

(c) a^{1 / 3} ÷ a^{-2 / 3} = a^{(1 / 3) – (- 2 / 3)}

We get,

= a^{(1 / 3) + (2 / 3)}

= a^{(1 + 2) / 3}

= a¹

= 0

(d) a^-3 × a² × a⁰ = a^{-3 + 2 + 0}

= a^-1

We get,

= (1 / a)

(e) (b^-2 – a^-2) ÷ (b^-1 – a^-1)

This can be written as,

= (1/b² – 1/a²) / (1/b – 1/a)

= {(1/b)² – (1/a)²} / (1/b – 1/a)

= {(1/b + 1/a) (1/b – 1/a)} / (1/b – 1/a)

We get,

= (1/b + 1/a)

4. Evaluate the following:

(i) (2³ × 3⁵ × 24²) / (12² × 18³ × 27)

(ii) (4³ × 3⁷ × 5⁶) / (5⁸ × 2⁷ × 3³)

(iii) (12² × 75^-2 × 35 × 400) / (48² × 15^-3 × 525)

(iv) (2⁶ × 5^-4 × 3^-3 × 4²) / (8³ × 15^-3 × 25^-1)

Solution:

(i) (2³ × 3⁵ × 24²) / (12² × 18³ × 27)

This can be written as,

= {2³ × 3⁵ × (2³ × 3)²} / (2² × 3)² × (2 × 3²)³ × (3³)

= (2³ × 3⁵ × 2⁶ × 3²) / (2⁴ × 3² × 2³ × 3⁶ × 3³)

= (2⁹ × 3⁷) / (2⁷ × 3¹¹)

On further calculation, we get,

= (2^{9 – 7} / 3^{11 – 7})

= 2² / 3⁴

We get,

= (4 / 81)

(ii) (4³ × 3⁷ × 5⁶) / (5⁸ × 2⁷ × 3³)

This can be written as,

= (2²)³ × 3^{7 – 3}} / (5^{8 – 6} × 2⁷)

= (2⁶ × 3^{7 – 3}) / (5^{8 – 6} × 2⁷)

On further calculation, we get,

= (2⁶ × 3⁴) / (5² × 2⁷)

= {3⁴ / (5² × 2^{7 – 6})}

= {81 / (5² × 2¹)}

We get,

= (81 / 50)

(iii) (12² × 75^-2 × 35 × 400) / (48² × 15^-3 × 525)

This can be written as,

= (12² × 35 × 400 × 15³) / (48² × 525 × 75²)

= {(2² × 3)² × (7 × 5) × (2⁴ × 5²) × (3 × 5)³} / {(2⁴ × 3)² × (3 × 5² × 7) × (3 × 5²)²}

On calculating further, we get,

= (2^{4 + 4} × 3^{2+ 3} × 5^{1 + 2 + 3} × 7) / (2⁸ × 3^{2 + 1 + 2} × 5^{4 + 2} × 7)

= (2⁸ × 3⁵ × 5⁶ × 7) / (2⁸ × 3⁵ × 5⁶ × 7)

We get,

= 1

(iv) (2⁶ × 5^-4 × 3^-3 × 4²) / (8³ × 15^-3 × 25^-1)

= (2⁶ × 4² × 15³ × 25¹) / (8³ × 5⁴ × 3³)

This can be written as,

= {(2⁶ × (2²)² × (3 × 5)³ × (5²)¹} / {(2³)³ × 5⁴ × 3³}

= (2^{6 + 4} × 3³ × 5^{3 + 2}) / (2⁹ × 3³ × 5⁴)

= (2^{10 – 9} × 5^{5 – 4})

= (2 × 5)

We get,

= 10

5. Simplify the following and express with positive index:

(a) 3p^-2q³ ÷ 2p³q^-2

(b) {(p^-3)^{2 / 3}}^{1 / 2}

Solution:

(a) 3p^-2q³ ÷ 2p³q^-2

This can be written as,

= (3p^-2q³) / (2p³q^-2)

= (3 / 2) {(p^-2 / p³) × (q³ / q^-2)}

= (3 / 2) {(p^-2 ÷ p³) × (q³ ÷ q^-2)}

= (3 / 2) {(p^{– 2 – 3}) × (q^{3 – (- 2)})} [Using a^m ÷ aⁿ = a^{m – n}]

= (3 / 2) {(p^-5) × (q⁵)}

= (3 / 2) {(1 / p⁵) × (q⁵)}

We get,

= (3q⁵ / 2p⁵)

(b) {(p^-3)^{2 / 3}}^{1 / 2}

= p^{-3 × (2 / 3) × (1 / 2)}

= p^{– 1}

We get,

= (1 / p)

6. Evaluate the following:

(i) {1 – (15 / 64)}^{– 1 / 2}

(ii) (8 / 27)^{– 2 / 3} – (1 / 3)^{– 2} – 7⁰

(iii) 9^{5 / 2} – 3 × 5⁰ – (1 / 81) ^{– 1 / 2}

(iv) (27)^{2 / 3} × (8)^{– 1 / 6} ÷ (18)^{– 1 / 2}

(v) (16)^{3 / 4} + 2 (1 / 2)^-1 × 3⁰

(vi)

Solution:

(i) {1 – (15 / 64)}^{– 1 / 2}

On taking LCM, we get,

= {(64 – 15) / 64}^{– 1 / 2}

= (49 / 64) ^{– 1 / 2}

= (64 / 49)^{1 / 2}

We get,

= (8 / 7)

(ii) (8 / 27)^{– 2 / 3} – (1 / 3)^{– 2} – 7⁰

= (27 / 8)^{2 / 3} – (1 / 3)^-2 – 7⁰

= (27 / 8)^{2 / 3} – (3)² – 1

On further calculation, we get,

= (3 / 2)^{3 × 2 / 3} – 9 – 1

= (3 / 2)² – 10

= (9 / 4) – 10

On taking LCM, we get,

= {(9 – 40) / 4}

= (- 31 / 4)

(iii) 9^{5 / 2} – 3 × 5⁰ – (1 / 81) ^{– 1 / 2}

On further calculation, we get,

= 3^{2 × 5 / 2} – 3 × 1 – (81)^{1 / 2}

= 3⁵ – 3 – 9^{2 × 1 / 2}

= 243 – 3 – 9

We get,

= 231

(iv) (27)^{2 / 3} × (8)^{– 1 / 6} ÷ (18)^{– 1 / 2}

This can be written as,

= 3^{3 × 2 / 3} × {1 / (2^{3 × 1 / 6})} ÷ (1 / 18)^{1 / 2}

= (3²) / (2^{1 / 2}) × (2 × 3²)^{1 / 2}

= (3² / 2^{1 / 2}) × 2^{1 / 2} × 3

We get,

= 3^{2 + 1}

= 3³

= 27

(v) (16)^{3 / 4} + 2 (1 / 2)^-1 × 3⁰

= 2^{4 × 3 / 4} + 2 × 2 × 1

On further calculation, we get,

= 2³ + 4

= 8 + 4

We get,

= 12

(vi)

= (1 / 2²)^{1 / 2} + (0.1)^-1 – 3²

= (1 / 2) + (0.1)^-1 – 3²

= (1 / 2) + (1 / 0.1) – 9

= (1 / 2) + (10 / 1) – 9

= (1 / 2) + 1

On taking LCM, we get,

= {(1 + 2) / 2}

= (3 / 2)

7. Simplify the following:

(a) (27x⁹)^{2 / 3}

(b) (8x⁶y³)^{2 / 3}

(c) (64a¹² / 27b⁶)^{– 2 / 3}

(d) (36m^-4 / 49n^-2)^{– 3 / 2}

(e) (a^{1 / 3} + a^{– 1 / 3}) (a^{2 / 3} – 1 + a^{– 2 / 3})

(f) ÷

(g) {(a^m)^{{m – (1 / m)}}}^{(1 / m + 1)}

(h) x^{m + 2n}. x^{3m – 8n} ÷ x^{5m – 60}

(i) (81)^{3 / 4} – (1 / 32)^{– 2 / 5} + 8^{1 / 3}. (1 / 2)^-1. 2⁰

(j) (27/343)^2/3 ÷ {1/ (625/1296)^1/4} x 536 /

Solution:

(a) (27x⁹)^{2 / 3}

This can be written as,

= (3³x⁹)^{2 / 3}

= (3³)^{2 / 3}(x⁹)^{2 / 3} [Using (a × b)ⁿ = aⁿ × bⁿ]

On calculating further,

We get,

= (3)^{3 × 2 / 3}(x)^{9 × 2 / 3} [Using (a^m)ⁿ = a^mn]

= (3)²x^{3 × 2}

We get,

= 9x⁶

(b) (8x⁶y³)^{2 / 3}

This can be written as,

= (2³x⁶y³)^{2 / 3}

= (2³)^{2 / 3} (x⁶)^{2 / 3}(y³)^{2 / 3} [Using (a × b)ⁿ = aⁿ × bⁿ]

= (2)^{3 × 2 / 3}(x)^{6 × 2 / 3}(y)^{3 × 2 / 3} [Using (a^m)ⁿ = a^mn]

= (2)²(x)^{2 × 2}(y)²

We get,

= 4x⁴y²

(c) (64a¹² / 27b⁶)^{– 2 / 3}

This can be written as,

= {(2⁶a¹²) / (3³b⁶)}^{– 2 / 3}

= {2^{6 × (- 2 / 3)}a^{12 × (- 2 / 3)}} / {3^{3 × (- 2 / 3)}b^{6 × (- 2 / 3)}}

On further calculation, we get,

= (2^{– 4}a^{– 8}) / (3^{– 2}b^{– 4})

= (3²b⁴) / (2⁴a⁸) [Using a^-n = (1 / aⁿ)]

We get,

= (9b⁴ / 16a⁸)

(d) (36m^-4 / 49n^-2)^{– 3 / 2}

This can be written as,

= {(6²m^-4) / (7²n^{– 2})}^{– 3 / 2}

= {6^{2 × (- 3 / 2)} m^{-4 × (- 3 / 2)}} / {7^{2 × (- 3 / 2)} n^{– 2 × (- 3 / 2)}}

On further calculation, we get,

= (6^-3 m⁶) / (7^{– 3}n³)

= (7³m⁶) / (6³n³)

We get,

= (343m⁶) / (216n³)

(e) (a^{1 / 3} + a^{– 1 / 3}) (a^{2 / 3} – 1 + a^{– 2 / 3})

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

On simplification, we get,

= (a^{1 / 3} × a^{2 / 3} – a^{1 / 3} × 1 + a^{1 / 3} × a^{– 2 / 3}) + (a^{– 1 / 3} × a^{2 / 3} – a^{– 1 / 3} × 1 + a^{– 1 / 3} × a^{– 2 / 3})

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

= {a¹ – a^{1 / 3} + a^{– 1 / 3}} + {a^{1 / 3} – a^{– 1 / 3} + a^{– 1}}

= {a – a^{1 / 3} + a^{– 1 / 3} + a^{1 / 3} – a^{– 1 / 3} + (1 / a)}

We get,

= {a + (1 / a)}

(f)
÷

This can be written as,

= (x⁴y²)^{1 / 3} ÷ (x⁵y^{– 5})^{1 / 6}

On calculating further, we get,

= {x^{4 × (1 / 3)} y^{2 × (1 / 3)}} ÷ {x^{5 × (1 / 6)} y^{– 5 × (1 / 6)}}

= {x^{(4 / 3)} y^{(2 / 3)}} ÷ {x^{(5 / 6)} y^{(- 5 / 6)}}

= {x^{(4 / 3)}y^{(2 / 3)}} / {x^{(5 / 6)} y^{(- 5 / 6)}}

= x^{(4 / 3) – (5 / 6)} y^{(2 / 3) – (- 5 / 6)} [Using a^m ÷ aⁿ = a^{m – n}]

= x^{(1 / 2)} y^{(3 / 2)}

= x^{(1 / 2)} (y³)^{(1 / 2)}

= √x √y³

We get,

=

(g) {(a^m)^{{m – (1 / m)}}}^{(1 / m + 1)}

= (a) ^{m × {m – (1 / m)} × {1 / (m + 1)}}

Now,

Consider,

m × {m – (1 / m)} × {1 / (m + 1)}

= (m² – 1) × {1 / (m + 1)}

= m² × {1 / (m + 1)} – 1 × {1 / (m + 1)}

= {m² / (m + 1)} – {1 / (m + 1)}

= {(m² – 1)} / {(m + 1)}

= {(m – 1) (m + 1)} / (m + 1)

We get,

= (m – 1)

Therefore, (a) ^{m × {m – (1 / m)} × {1 / (m + 1)}} = a^{m – 1}

(h) x^{m + 2n}. x^{3m – 8n} ÷ x^{5m – 60}

= x^{m + 2n + 3m – 8n – 5m – (- 60)}

= x^{m + 2n + 3m – 8n – 5m + 60}

We get,

= x^{– m – 6n + 60}

(i) (81)^{3 / 4} – (1 / 32)^{– 2 / 5} + 8^{1 / 3}. (1 / 2)^-1. 2⁰

This can be written as,

= (3⁴)^{(3 / 4)} – (1 / 2⁵)^{(- 2 / 5)} + (2³)^{(1 / 3)}. (1 / 2)^{– 1} × 1

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

= 3³ – {1 / 2^{– 2}} + 2¹. (2)¹

= 3³ – 2² + 2^{(1 + 1)}

= 3³ – 2² + 2²

We get,

= 3³

= 27

(j) (27/343)^2/3 ÷ {1/ (625/1296)^1/4} x 536 /

This can be written as,

= {3³ / (7³}^2/3 ÷ [1/{5⁴ / (2⁴ x 3⁴)}^1/4] x (2³ x 67) /

= (3³/ 7³)^2/3 ÷ [1/ {5⁴ / (2⁴ x 3⁴)}^1/4] x (2³ x 67) / (3³)^1/3

= (3^{3 x 2/3} / 7^{3 x 2/3}) ÷ [1/ {5^{4 x ¼} / (2^{4 x ¼} x 3^{4 x ¼})] x (2³ x 67) / (3^{3 x 1/ 3})

Using (a^m)ⁿ = a^{m n}

= (3² / 7²) ÷ [1 / {5¹ / (2¹ x 3¹)}] x (2³ x 67) / 3¹

= (3² / 7²) ÷ {(2¹ x 3¹) / 5¹} x (2³ x 67) / 3¹

= (3² / 7²) x {5¹ / (2¹ x 3¹)} x (2³ x 67) / 3¹

On further calculation, we get,

= 3^{2 – 1 – 1} x 2^{3 – 1} x 5¹ x 7^-2 x 67

Using a a^m ÷ aⁿ = a^{m – n}

= 3⁰ x 2² x 5¹ x 7^-2 x 67

= 1 x 4 x 5¹x 7^-2 x 67

= 1340 / 49

= 27.34

8. Simplify the following:

(i) (5^x × 7 – 5^x) / (5^{x + 2} – 5^{x + 1})

(ii) (3^{x + 1} + 3^x) / (3^{x + 3} – 3^{x + 1})

(iii) (2^m × 3 – 2^m) / (2^{m + 4} – 2^{m + 1})

(iv) (5^{n + 2} – 6.5^{n + 1}) / (13. 5ⁿ – 2. 5^{n + 1})

Solution:

(i) (5^x × 7 – 5^x) / (5^{x + 2} – 5^{x + 1})

On taking common terms, we get,

= {5^x (7 – 1)} / {5^{x + 1} (5 – 1)}

= (5^{x – x – 1} × 6) / 4 [Using a^m ÷ aⁿ = a^{m – n}]

= (5^{– 1} × 6) / 4

= {6 / (5 × 4)}

We get,

= (3 / 10)

(ii) (3^{x + 1} + 3^x) / (3^{x + 3} – 3^{x + 1})

On taking common terms, we get,

= {3^x (3 + 1)} / {3^x (3³ – 3)}

= {4 / (27 – 3)}

= (4 / 24)

We get,

= (1 / 6)

(iii) (2^m × 3 – 2^m) / (2^{m + 4} – 2^{m + 1})

On taking common terms, we get,

= {2^m (3 – 1)} / {2^m (2⁴ – 2)}

= {2 / (16 – 2)}

= (2 / 14)

We get,

= (1 / 7)

(iv) (5^{n + 2} – 6.5^{n + 1}) / (13. 5ⁿ – 2. 5^{n + 1})

On taking common terms, we get,

= {5ⁿ (5² – 6 × 5)} / {5ⁿ (13 – 2 × 5)}

= (25 – 30) / (13 – 10)

We get,

= (- 5 / 3)

9. Solve for x:

(a) 2^{2x + 1} = 8

(b) 3 × 7^x = 7 × 3^x

(c) 2^{x + 3} + 2^{x + 1} = 320

(d) 9 × 3^x = (27)^{2x – 5}

(e) 2^{2x + 3} – 9 × 2^x + 1 = 0

(f) 1 = p^x

(g) p³ × p^{– 2} = p^x

(h) p^{– 5} = (1 / p^{x + 1})

(i) 2^2x + 2^{x + 2} – 4 × 2³ = 0

(j) 9 x 81^x = 1/ 27^{x – 3}

(k) 2^{2x – 1} – 9 × 2^{x – 2} + 1 = 0

(l) : 5^x = 25: 1

(m) = (0.6)^{2 – 3x}

(n) = (27^{– 1}) / (125^{– 1})

(o) 9^{x + 4} = 3² × (27)^{x + 1}

Solution:

(a) 2^{2x + 1} = 8

This can be written as,

2^{2x + 1} = 2³

2x + 1 = 3

2x = 3 – 1

2x = 2

We get,

x = 1

(b) 3 × 7^x = 7 × 3^x

(7^x / 7) = (3^x / 3)

7^{x – 1} = 3^{x – 1} [Using a^m ÷ aⁿ = a^{m – n}]

7^{x – 1}= 3^{x – 1} × 1

7^{x – 1} = 3^{x – 1} × 7⁰

x – 1 = 0

We get,

x = 1

(c) 2^{x + 3} + 2^{x + 1} = 320

This can be written as,

2^{x + 3} + 2^{x + 1} = 2⁶ × 5

2^x. 2³ + 2^x.2¹ = 2⁶ × 5

On taking common terms, we get,

2^x (2³ + 2¹) = 2⁶ × 5

2^x (8 + 2) = 2⁶ × 5

2^x (10) = 2⁶ × 5

2^x (10 / 5) = 2⁶

2^x. 2 = 2⁶

(2^x.2) / 2⁶ = 1

2^{x + 1 – 6} = 1 × 2⁰

2^{x – 5} = 1 × 2⁰

x – 5 = 0

We get,

x = 5

(d) 9 × 3^x = (27)^{2x – 5}

This can be written as,

3² × 3^x = (3³)^{2x – 5}

3² × 3^x = 3^{3 × (2x – 5)}

On further calculation, we get,

3^{2 + x} = 3^{6x – 15}

1 = (3^{6x – 15}) / (3^{2 + x})

1 = 3^{6x – 15 – 2 – x}

3⁰ = 3^{5x – 17}

5x – 17 = 0

5x = 17

We get,

x = (17 / 5)

(e) 2^{2x + 3} – 9 × 2^x + 1 = 0

This can be written as,

2^2x.2³ – 9 × 2^x + 1 = 0

Put 2^x = t

Then,

2^2x = t²

So,

2^2x.2³ – 9 × 2^x + 1 = 0 becomes,

8t² – 9t + 1 = 0

8t² – 8t – 1t + 1 = 0

On taking common terms, we get,

8t (t – 1) – 1 (t – 1) = 0

(t – 1) = 0 or (8t – 1) = 0

t = 1 or t = (1 / 8)

2^x = 1 or 2^x = (1 / 2³)

2^x = 2⁰ or 2^x = 2^-3

Hence,

x = 0 or x = – 3

(f) 1 = p^x

p⁰ = p^x [Using a⁰ = 1]

Therefore,

x = 0

(g) p³ × p^{– 2} = p^x

p^{3 + (- 2)} = p^x [Using a^m × aⁿ = a^{m + n}]

p^{3 – 2} = p^x

p¹ = p^x

Hence,

x = 1

(h) p^{– 5} = (1 / p^{x + 1})

p^-5 × p^{x + 1} = 1

p^{– 5 + x + 1} = 1 [Using a^m × aⁿ = a^{m + n}]

p^{x – 4} = p⁰

x – 4 = 0

We get,

x = 4

(i) 2^2x + 2^{x + 2} – 4 × 2³ = 0

This can be written as,

2^2x + 2^{x + 2} – 2² × 2³ = 0

2^2x + 2^x.2² – 2^{2 + 3} = 0 [Using a^m × aⁿ = a^{m + n}]

2^2x + 2^x.2² – 2⁵ = 0

2^2x + 2^x. 4 – 32 = 0

Put 2^x = t

Then, 2^2x = t²

2^2x + 2^x.4 – 32 = 0 becomes,

t² + 4t – 32 = 0

t² + 8t – 4t – 32 = 0

On taking common terms, we get,

t (t + 8) – 4 (t + 8) = 0

t + 8 = 0 or t – 4 = 0

t = – 8 or t = 4

2^x = – 8 or 2^x = 4

2^x = – 2³ or 2^x = 2²

Now,

Consider second equation,

2^x = 2²

We get,

x = 2

(j) 9 x 81^x = 1/ 27^{x – 3}

This can be written as,

3² x 3^4x = 1/ 3^{3 (x – 3)}

3² x 3^4x = 1/ 3^{3x – 9}

Using (a^m)ⁿ = a^mn

3² x 3^4x x 3^{3x – 9} = 1

3^{2 + 4x + 3x – 9} = 1 x 3⁰

On further calculation, we get,

2 + 4x + 3x – 9 = 0

7x – 7 = 0

7x = 7

x = 1

(k) 2^{2x – 1} – 9 × 2^{x – 2} + 1 = 0

This can be written as,

2^2x. 2^-1 – 9 × 2^x. 2^{– 2} + 1 = 0

Let 2^x = t,

So, 2^2x = t²

Then,

2^2x.2^{– 1} – 9 × 2^x.2^{– 2} + 1 = 0 becomes,

(t² / 2) – 9 × (t / 2²) + 1 = 0

(t² / 2) – (9t / 4) + 1 = 0

Taking LCM, we get,

2t² – 9t + 4 = 0

2t² – 8t – t + 4 = 0

2t (t – 4) – 1 (t – 4) = 0

t – 4 = 0 or 2t – 1 = 0

t = 4 or t = (1 / 2)

Hence,

2^x = 4 or 2^x = (1 / 2)

2^x = 2² or 2^x = 2^{– 1}

Therefore,

x = 2 or x = – 1

(l)
: 5^x = 25: 1

This can be written as,

(
/ 5^x) = (25 / 1)

(
/ 5^x) = (5² / 1)

= 5² × 5^x

= 5^{2 + x}

x² = 2 + x

x² – x – 2 = 0

(x – 2) (x + 1) = 0

x – 2 = 0 or x + 1 = 0

Therefore,

x = 2 or x = – 1

(m)
= (0.6)^{2 – 3x}

This can be written as,

{1 + (2 / 3)}^{1 / 2} = (6 / 10)^{2 – 3x}

On taking LCM, we get,

(5 / 3)^{1 / 2} = (3 / 5)^{2 – 3x}

(3 / 5)^{– 1 / 2} = (3 / 5)^{2 – 3x}

(- 1 / 2) = 2 – 3x

-1 = 2 (2 – 3x)

– 1 = 4 – 6x

– 1 – 4 = – 6x

– 5 = – 6x

Hence,

x = (5 / 6)

(n)
= (27^{– 1}) / (125^{– 1})

This can be written as,

(3 / 5)^{(x + 3) × (1 / 2)} = {(3³)^-1 / (5³)^-1}

(3 / 5)^{(x + 3) / 2} = (3 / 5)^{– 3}

(x + 3) / 2 = – 3

x + 3 = – 3 × 2

x + 3 = – 6

x = – 6 – 3

We get,

x = – 9

(o) 9^{x + 4} = 3² × (27)^{x + 1}

This can be written as,

9^{x + 4} = 3² × (3³)^{x + 1}

3^{2 (x + 4)} = 3² × 3^{3x + 3}

3^{2x + 8} = 3^{2 + 3x + 3}

Hence,

2x + 8 = 2 + 3x + 3

2x + 8 = 3x + 5

3x – 2x = 8 – 5

We get,

x = 3

10. Find the value of k in each of the following:

(i) = 2^k

(ii) = x^k

(iii) × = 3^k

(iv) (1 / 3)^{– 4} ÷ 9 ^{(- 1 / 3)} = 3^k

Solution:

(i)
= 2^k

This can be written as,

8^{(1 / 3) × (- 1 / 2)} = 2^k

(2³)^{(1 / 3) × (- 1 / 2)} = 2^k

2^{(- 1 / 2)} = 2^k

Therefore,

k = (- 1 / 2)

(ii)
= x^k

This can be written as,

{(x²)^{1 / 3}}^{1 / 4} = x^k

On further calculation, we get,

(x²)^{1 / 12} = x^k

x^{(2 / 12)} = x^k

x^{(1 / 6)} = x^k

Hence,

k = (1 / 6)

(iii)
×
= 3^k

This can be written as,

{(3²)^{(1 / 2)}}^{– 7} × {(3)^{(1 / 2)}}^{– 5} = 3^k

3^{– 7} × 3^{(- 5 / 2)} = 3^k

3^{-7 – 5 / 2} = 3^k

3^{(- 14 – 5) / 2} = 3^k

3^{(- 19 / 2)} = 3^k

Therefore,

k = (- 19 / 2)

(iv) (1 / 3)^{– 4} ÷ 9 ^{(- 1 / 3)} = 3^k

This can be written as,

(3^{– 1})^{– 4} ÷ (3²)^{– 1 / 3} = 3^k

3⁴ ÷ 3^{(- 2 / 3)} = 3^k

3^{4 – (- 2 / 3)} = 3^k

3^{4 + 2 / 3} = 3^k

3^{(14 / 3)} = 3^k

We get,

k = (14 / 3)

11. If a = 2^{(1 / 3)} – 2^{(- 1 / 3)}, prove that 2a³ + 6a = 3

Solution:

Given

a = 2^{(1 / 3)} – 2^{(- 1 / 3)}

This can be written as,

a = 2^{(1 / 3)} – {1 / 2^{(1 / 3)}}

On taking cube on both sides, we get,

a³ = [2^{(1 / 3)} – {1 / 2^{(1 / 3)}}]³

On further calculation, we get,

a³ = 2 – (1 / 2) – 3 [2^{(1 / 3)} – {1 / 2^{(1 / 3)}}]

a³ = {(4 – 1) / 2} – 3a

a³ = (3 / 2) – 3a

We get,

2a³ + 6a = 3

12. If x = 3^{(2 / 3)} + 3^{(1 / 3)}, prove that x³ – 9x – 12 = 0

Solution:

Given

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

x³ = 3² + 3 + 3 × 3^{2 / 3} × 3^{1 / 3} (3^{2 / 3} + 3^{1 / 3})

x³ = 9 + 3 + 3 × 3^{2 / 3 + 1 / 3} (x)

x³ = 12 + 9x

We get,

x³ – 9x – 12 = 0

13. If and abc = 1, prove that x + y + z = 0

Solution:

Let
= k

So,

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

We get,

a = k^x, b = k^y , c = k^z

Also, given that,

abc = 1

k^x × k^y × k^z = 1

k^{x + y + z} = k⁰

We get,

x + y + z = 0

14. If a^x = b^y = c^z and b² = ac, prove that y = 2xz / (z + x).

Solution:

Let a^x = b^y = c^z = k

So,

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

Also given that,

b² = ac

k^{2 / y} = k^{1 / x} × k^{1 / z}

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

(2 / y) = (1 / x) + (1 / z)

(2 / y) = (z + x) / zx

We get,

y = {2zx / (z + x)}

15. Show that: {1 / (1 + a^{p – q})} + {1 / (1 + a^{q – p})} = 1

Solution:

Consider LHS of the equation, i.e,

{1 / (1 + a^{p – q})} + {1 / (1 + a^{q – p})}

On taking LCM, we get,

= {(1 + a^{q – p}) + (1 + a^{p – q})} / (1 + a^{p – q}) (1 + a^{q – p})

= (2 + a^{– (p – q)} + a^{p – q}) / (1 + a^{p – q}) (1 + a^{– (p – q)})

= 2 + a^{– (p – q)} + a^{p – q} / (1 + a^{– (p – q)} + a^{p – q} + a^{p – q}. a^{– (p – q)})

= 2 + a^{– (p – q)} + a^{p – q} / (1 + a^{– (p – q)} + a^{p – q} + a^{p – q – p + q}

= 2 + a^{– (p – q)} + a^{p – q} / (1 + a^{– (p – q)} + a^{p – q} + a⁰)

= 2 + a^{– (p – q)} + a^{p – q} / (1 + a^{– (p – q)} + a^{p – q} + 1)

= 2 + a^{– (p – q)} + a^{p – q} / (2 + a^{– (p – q)} + a^{p – q})

We get,

= 1

= RHS

Therefore,

LHS = RHS

Hence, proved