Frank Chapter 9 Indices ICSE Solutions Class 9 Math

Frank Solutions for Chapter 9 Indices Class 9 Mathematics ICSE

Exercise 9.1

1. Evaluate the following:

(i) 6⁰

(ii) (1/2)^-3

(iii) 2³²

(iv) 3²³

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

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

(vii)

(viii)

Answer

(i) 6⁰ = 1

(ii) (1/2)^-3 = (2)³

We get,

(1/2)^-3 = 8

(iii) = 2⁶

We get,

= 64

(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)

(viii)

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)

Answer

(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 × (-32)/(6)^{2 × (- 3/2)}}

[Using (a^m)ⁿ = a^mn]

= {(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}

[Using a^-m = (1/a^m)

= {(2)⁴/(2)⁶} × {(6)³/(6)²}

= (2)^{4 – 6} × (6)^{3 – 2}

[Using a^m ÷ aⁿ = a^{m – n}]

= (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)

Answer

(a) (a³)⁵ × a⁴ = (a)^3×5 × a⁴

[Using (a^m)ⁿ = a^mn]

we get,

= (a)¹⁵ × a⁴

= a^{15 + 4}

[Using a^m × aⁿ = a^{m + n}]

We get,

= a¹⁹

(b) a² × a³ ÷ a⁴ = a^{2 + 3 – 4}

[Using a^m × aⁿ = a^{m + n} and a^m ÷ aⁿ = a^{m – n}]

We get,

= a¹

= a

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

[Using a^m ÷ aⁿ = a^{m – n}]

We get,

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

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

= a¹

= 0

(d) a^-3 × a² × a⁰ = a^{-3 + 2 + 0}

[Using a^m × aⁿ = a^{m + n}]

= 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)

Answer

(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

Answer

(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)}

[Using (a^m)ⁿ = a^mn]

= 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) 

Answer

(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)

Answer

(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)}

[Using (a×b)ⁿ = aⁿ×bⁿ and (a/b)ⁿ = (aⁿ/bⁿ)]

On further calculation, we get,

= (2^–4a^–8)/(3^–2b^–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)}

[Using (a×b)ⁿ = aⁿ × bⁿ and (a/b)ⁿ = (aⁿ/bⁿ)]

On further calculation, we get,

= (6^-3 m⁶)/(7^{– 3}n³)

= (7³m⁶)/(6³n³)

[Using a^–1 = (1/aⁿ)]

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)}}

[Using a^m × aⁿ = a^m+n]

= {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)}

[Using (a^m)ⁿ = a^mn]

= {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)

[Using (a^m)ⁿ = a^mn]

= √x √y³

We get,

(g) {(a^m)^{m–(1/m)}}^{(1/m +1)}

= (a)^{m×{m–(1/m)}×{1/(m+1)}}

[Using (a^m)ⁿ = a^mn]

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)}

[Using a^m×aⁿ = a^m+n and a^m ÷ aⁿ = a^{m – n}]

= 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

[Using a⁰ = 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⁾

[Using a^m × aⁿ = a^{m + n} and a^{– n} = (1/aⁿ)]

= 3³ – 2² + 2²

We get,

= 3³

= 27

(j) 

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) (5ⁿ⁺² – 6.5ⁿ⁺¹)/(13.5ⁿ – 2.5ⁿ⁺¹)

Answer

(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) (5ⁿ⁺² – 6.5ⁿ⁺¹)/(13.5ⁿ – 2.5ⁿ⁺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^xz:5^x= 25:1

(m)

(n) 

(o) 9^{x + 4} = 3² × (27)^{x + 1}

Answer

(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 [Using a⁰ = 1]

⇒ 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 – 2²⁺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) 9×81^x = 1/27^{x – 3}

This can be written as,

3²×3^4x = 1/3^{3 (x – 3)}

⇒ 3²×3^4x = 1/3^{3x – 9}

Using (a^m)ⁿ = a^mn

3² ×3^4x ×3^3x–9 = 1

⇒ 3^2+4x+3x–9 = 1×3⁰

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)

Therefore,

x = 2 or x = – 1

(m)

(n)

(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)

(ii)

(iii)

(iv) (1/3)^{– 4} ÷ 9 ^{(- 1/3)} = 3^k

Answer

(i)

(ii)

(iii)

(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

Answer

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

Answer

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. Ifand abc = 1, prove that x + y + z = 0

Answer

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).

Answer

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

Answer

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

16. Find the value of (8p)^P if 9^p+2 – 9^p = 240.

Answer

17. If a^x = b^y = c^z and abc = 1, show that 1/x + 1y + 1/z = 0.

Answer

18. If x ^1/3 + y ^1/3 + z ^1/3 = 0, prove that (x + y + z)³ = 27xyz

Answer

19. If 2250 = 2^a .3^b. 5^c, find a, b and c. Hence, calculated the value of 3^a × 2^-b × 5^-c

Answer

20. If 2400 = 2^x × 3^y × 5^z, find the numerical value of x, y, z. Find the value of 2^-x × 3^y × 5^z as fraction.

Answer

21. If 2^x = 3^y = 12^z; show that 1/z = 1/y + 2/x.

Answer

22. (a) Find the value of ‘a’ and ‘b’ if:

(b) Find the value of ‘a’ and ‘b’ if:

Answer

(a)

(b)

23. (a) Prove the following:

 
(b) Prove the following:

(c) Prove the following:

(d) Prove the following:

Answer

(a)

(b)

(c)

(d)