ML Aggarwal Logarithms Exe-9.2 Class 9 ICSE Maths Solutions

ML Aggarwal Logarithms Exe-9.2 Class 9 ICSE Maths Solutions Chapter-9. Step by Step solutions of Logarithms ML Aggarwal ICSE 9th.. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Logarithms Exe-9.2 Class 9 ICSE Maths Solutions Chapter-9

Board	ICSE
Subject	Maths
Class	9th
Chapter-9	Logarithms
Topics	Solution of Exe-9.2 Questions
Academic Session	2024-2025

Solution of Exe-9.2 Questions on Logarithms

ML Aggarwal Logarithms Exe-9.2 Class 9 ICSE Maths Solutions Chapter-9

Question 1. Simplify the following:

(i) log a³ – log a²

(ii) log a³ ÷ log a²

(iii) log 4/log 2

(iv) (log 8 log 9)/log 27

(v) log 27/log √3

(vi) (log 9 – log 3)/log 27

Answer:

(i) log a³ – log a²

By using Quotient law,

log a³ – log a² = log (a³/a²)

= log a

(ii) log a³ ÷ log a²

By using power law,

log a³ ÷ log a² = 3log a ÷ 2 log a

= 3log a / 2log a

= 3/2

(iii) log 4/log 2

Let us simplify the expression,

log 4/log 2 = log(2×2)/log 2

By using power law,

= 2 log 2/log 2

= 2

(iv) (log 8 log 9)/log 27

Let us simplify the expression,

(log 8 log 9)/log 27 = (log 2³. log 3²)/log 3³

By using power law,

= [(3 log 2).(2 log 3)]/(3 log 3)

= [(log 2).2] / 1

= 2 log 2

= log 2²

= log 4

(v) log 27/log √3

Let us simplify the expression,

log 27/log √3 = log(3×3×3)/log(3)^1/2

= log 3³/log 3^1/2

By using power law

= 3log 3/((1/2)log 3)

= (3×2)/1 (log 3/log 3)

= (6) (1)

= 6

(vi) (log 9 – log 3)/log 27

Let us simplify the expression,

(log 9 – log 3)/log 27 = [log(3×3) – log 3] / log(3×3×3)

= [log 3² – log 3] / log 3³

By using power law

= [2 log 3 – log 3]/3log 3

= log 3/3log 3

= 1/3

Question 2. Evaluate the following:

(i) log (10 ÷ ∛10)

(ii) 2 + ½ log(10^-3)

(iii) 2 log 5 + log 8 – ½ log 4

(iv) 2 log 10³ + 3 log 10^-2 – 1/3 log 5^-3 + ½ log 4

(v) 2 log 2 + log 5 – ½ log 36 – log 1/30

(vi) 2 log 5 + log 3 + 3 log 2 – ½ log 36 – 2 log 10

(vii) log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80

(viii) 2 log₁₀ 5 + log₁₀ 8 – ½ log₁₀ 4

Answer:

(i) log (10 ÷ ∛10)

Let us simplify the expression,

log (10 ÷ ∛10) = log (10 ÷ 10^1/3)

= log (10^{1 – 1/3})

= log (10^2/3)

= 2/3 log 10

= 2/3 (1)

= 2/3

(ii) 2 + ½ log(10^-3)

Let us simplify the expression,

2 + ½ log(10^-3) = 2 + ½ × (-3) log 10

= 2 – 3/2 log 10

= 2 – 3/2 (1)

= 2 – 3/2

= (4-3)/2

= ½

(iii) 2 log 5 + log 8 – ½ log 4

Let us simplify the expression,

2 log 5 + log 8 – ½ log 4 = log 5² + log 8 – ½ log 2²

= log 25 + log 8 – ½ 2log 2

= log 25 + log 8 – log 2

= log (25×8)/2

= log (25×4)

= log 100

= log 10²

= 2 log 10

= 2 (1)

= 2

(iv) 2 log 10³ + 3 log 10^-2 – 1/3 log 5^-3 + ½ log 4

Let us simplify the expression,

2 log 10³ + 3 log 10^-2 – 1/3 log 5^-3 + ½ log 4 = 2×3 log 10 + 3(-2)log 10 – 1/3 (-3) log 5 + ½ log 2²

= 6 log 10 – 6 log 10 + log 5 + ½ 2 log 2

= 6 log 10 – 6 log 10 + log 5 + log 2

= 0 + log 5 + log 2

= log (5×2)

= log 10

= 1

(v) 2 log 2 + log 5 – ½ log 36 – log 1/30

Let us simplify the expression,

2 log 2 + log 5 – ½ log 36 – log 1/30 = log 2² + log 5 – ½ log 6² – log (1/30)

= log 4 + log 5 – log 6 – log 1/30

= log 4 + log 5 – log 6 – (log 1 – log 30)

= log 4 + log 5 – log 6 – log 1 + log 30

= log 4 + log 5 + log 30 – (log 6 + log 1)

= log (4×5×30) – log (6×1)

= log (4×5×30)/(6×1)

= log (4×5×5)

= log 100

= log 10²

= 2 log 10

= 2 (1)

= 2

(vi) 2 log 5 + log 3 + 3 log 2 – ½ log 36 – 2 log 10

Let us simplify the expression,

2log 5 + log 3 + 3log 2 – ½ log 36 – 2log 10 = log 5² + log 3 + log 2³ – ½ log 6² – log 10²

= log 25 + log 3 + log 8 – log 6 – log 100

= log (25×3×8) – log(6×100)

= log (25×3×8)/ (6×100)

= log (1×3×8) / (6×4)

= log 24/24

= log 1

= 0

(vii) log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80

Let us simplify the expression,

log 2 + 16 log 16/15 + 12 log 25/24 + 7 log 81/80 = log 2 + 16(log 16 – log 15) + 12(log 25 – log 24) + 7(log 81 – log 80)

= log 2 + 16 (log 2⁴ – log (3×5)) + 12 (log 5² – log (3×2×2×2)) + 7 (log (3×3×3×3) – log (2⁴ × 5))

= log 2 + 16(4log 2 – (log 3 + log 5)) + 12 (2log 5 – log (3×2³)) + 7 (log 3⁴ – (log 2⁴ + log 5))

= log 2 + 16 (4log 2 – log 3 – log 5) + 12 (2log 5 – (log 3 + 3log 2)) + 7 (4log 3 – 4log 2 – log 5)

= log 2 + 64log 2 – 16log 3 – 16log 5 + 24log 5 – 12log 3 – 36log 2 + 28log 3 – 28log 2 – 7log 5

= (log 2 + 64log 2 – 36log 2 – 28log 2) + (-16log 3 – 12log 3 + 28log 3) + (-16log 5 + 24log 5 – 7log 5)

= (65log 2 – 64log 2) + (-28log 3 + 28log 3) + (-23log 5 + 24log 5)

= log 2 + 0 + log 5

= log (2×5)

= log 10

= 1

(viii) 2 log₁₀ 5 + log₁₀ 8 – ½ log₁₀ 4

Let us simplify the expression,

2 log₁₀ 5 + log₁₀ 8 – ½ log₁₀ 4 = log₁₀ 5² + log₁₀ 8 – log₁₀ 4^1/2

= log₁₀ 25 + log₁₀ 8 – log₁₀ (2)²×^1/2

= log₁₀ 25 + log₁₀ 8 – log₁₀ 2

= log₁₀ [(25×8)/2]

= log₁₀ (25×4)

= log₁₀ 100

= log₁₀ 10²

= 2 log₁₀ 10

= 2 (1)

= 2

Question 3. Express each of the following as a single logarithm:

(i) 2 log 3 – ½ log 16 + log 12

(ii) 2 log₁₀ 5 – log₁₀ 2 + 3 log₁₀ 4 + 1

(iii) ½ log 36 + 2 log 8 – log 1.5

(iv) ½ log 25 – 2 log 3 + 1

(v) ½ log 9 + 2 log 3 – log 6 + log 2 – 2

Answer:

(i) 2 log 3 – ½ log 16 + log 12

Let us simplify the expression into single logarithm,

2 log 3 – ½ log 16 + log 12 = 2 log 3 – ½ log 4² + log 12

= 2 log 3 – log 4 + log 12

= log 3² – log 4 + log 12

= log 9 – log 4 + log 12

= log (9×12)/4

= log (9×3)

= log 27

(ii) 2 log₁₀ 5 – log₁₀ 2 + 3 log₁₀ 4 + 1

Let us simplify the expression into single logarithm,

2 log₁₀ 5 – log₁₀ 2 + 3 log₁₀ 4 + 1 = log₁₀ 5² – log₁₀ 2 + log₁₀ 4³ + log₁₀ 10

= log₁₀ 25 – log₁₀ 2 + log₁₀ 64 + log₁₀ 10

= log₁₀ (25×64×10) – log₁₀ 2

= log₁₀ (16000) – log₁₀ 2

= log₁₀ (16000/2)

= log₁₀ 8000

(iii) ½ log 36 + 2 log 8 – log 1.5

Let us simplify the expression into single logarithm,

½ log 36 + 2 log 8 – log 1.5 = log 36^1/2 + log 8² – log 1.5

= log (6)²×^1/2 + log 64 – log 1.5

= log 6 + log 64 – log (15/10)

= log 6 + log 64 – (log 15 – log 10)

= log (6×64) – log 15 + log 10

= log (6×64×10) – log 15

= log [(6×64×10)/15]

= log (4×64)

= log 256

(iv) ½ log 25 – 2 log 3 + 1

Let us simplify the expression into single logarithm,

½ log 25 – 2 log 3 + 1 = log 25^1/2 – log 3² + log 10

= log (5)²×^1/2 – log 9 + log 10

= log 5 – log 9 + log 10

= log (5×10) – log 9

= log ((5×10)/9)

= log 50/9

(v) ½ log 9 + 2 log 3 – log 6 + log 2 – 2

Let us simplify the expression into single logarithm,

½ log 9 + 2 log 3 – log 6 + log 2 – 2 = log 9^1/2 + log 3² – log 6 + log 2 – log 100

= log 3²×^1/2 + log 9 – log 6 + log 2 – log 100

= log 3 + log 9 – log 6 + log 2 – log 100

= log [(3×9×2)/(6×100)]

= log 9/100

Question 4. Prove the following:

(i) log₁₀ 4 ÷ log₁₀ 2 = log₃ 9

(ii) log₁₀ 25 + log₁₀ 4 = log₅ 25

Answer:

(i) log₁₀ 4 ÷ log₁₀ 2 = log₃ 9

Let us consider LHS, log₁₀ 4 ÷ log₁₀ 2

log₁₀ 4 ÷ log₁₀ 2 = log₁₀ 2² ÷ log₁₀ 2

= 2 log₁₀ 2 ÷ log₁₀ 2

= 2 log₁₀ 2/ log₁₀ 2

= 2 (1)

= 2

Now let us consider RHS,

log₃ 9 = log₃ 3²

= 2 log₃ 3

= 2(1)

= 2

∴ LHS = RHS

Hence proved.

(ii) log₁₀ 25 + log₁₀ 4 = log₅ 25

Let us consider LHS, log₁₀ 25 + log₁₀ 4

log₁₀ 25 + log₁₀ 4 = log₁₀ (25×4)

= log₁₀ 100

= log₁₀ 10²

= 2 log₁₀ 10

= 2(1)

= 2

Now, let us consider RHS,

log₅ 25 = log₅ 5²

= 2 log₅ 5

= 2 (1)

= 2

∴ LHS = RHS

Hence proved.

Question 5. If x = (100)^a, y = (10000)^b and z = (10)^c, express log [(10√y)/x²z³] in terms of a, b, c.

Answer:

Given:

x = (100)^a = (10²)^a = 10^2a

y = (10000)^b = (10⁴)^b = 10^4b

z = (10)^c

It is given that, log [(10√y)/x²z³]

log [(10√y)/x²z³] = (log 10 + log √y) – (log x² + log z³)

= (1 + log(y)^1/2) – (log x² + log z³) [we know that, log 10 = 1]

= (1 + ½ log y) – (2 log x + 3 log z)

Now substitute the values of x, y, z, we get

= (1 + ½ log 10^4b) – (2 log 10^2a + 3 log 10^c)

= (1 + ½ 4b log 10) – (2×2a log 10 + 3×c log 10)

= (1 + ½ 4b) – (2×2a + 3c) [Since, log 10 = 1]

= (1 + 2b) – (4a + 3c)

= 1 + 2b – 4a – 3c

Question 6. If a = log₁₀x, find the following in terms of a :

(i) x

(ii) log₁₀ ⁵√x²

(iii) log₁₀ 5x

Answer:

Given:

a = log₁₀x

(i) x

10^a = x

∴ x = 10^a

(ii) log₁₀ ⁵√x²

log₁₀ ⁵√x² = log₁₀ (x²)^1/5

= log₁₀ (x)^2/5

= 2/5 log₁₀ x

= 2/5 (a)

= 2a/5

(iii) log₁₀ 5x

x = (10)^a

= log₁₀ 5x

= log₁₀ 5(10)^a

= log₁₀ 5 + log₁₀ 10

= log₁₀ 5 + a(1)

= a + log₁₀ 5

Question 7. If a =log 2/3, b = log 3/5 and c = 2 log √(5/2). Find the value of

(i) a + b + c

(ii) 5^a+b+c

Answer:

Given:

a =log 2/3

b = log 3/5

c = 2 log √(5/2)

(i) a + b + c

Let us substitute the given values, we get

a + b + c = log 2/3 + log 3/5 + 2 log √(5/2)

= (log 2 – log 3) + (log 3 – log 5) + 2 log (5/2)^1/2

= log 2 – log 3 + log 3 – log 5 + 2 × ½ (log 5 – log 2)

= log 2 – log 3 + log 3 – log 5 + log 5 – log 2

= 0

(ii) 5^a+b+c

5^a+b+c = 5⁰

= 1

Question 8. If x = log 3/5, y = log 5/4 and z = 2 log √3/2, find the value of

(i) x + y – z

(ii) 3^x+y-z

Answer:

Given:

x = log 3/5 = log 3 – log 5

y = log 5/4 = log 5 – log 4

z = 2 log √3/2 = log (√3/2)² = log ¾ = log 3 – log 4

(i) x + y – z

Let us substitute the given values, we get

x + y – z = log 3 – log 5 + log 5 – log 4 – (log 3 – log 4)

= log 3 – log 5 + log 5 – log 4 – log 3 + log 4

= 0

(ii) 3^x+y-z

3^x+y-z = 3⁰

= 1

Question 9. If x = log₁₀ 12, y = log₄ 2 × log₁₀ 9 and z = log₁₀ 0.4, find the values of

(i) x – y – z

(ii) 7^x-y-z

Answer:

Given:

x = log₁₀ 12

y = log₄ 2 × log₁₀ 9

z = log₁₀ 0.4

(i) x – y – z

Let us substitute the given values, we get

x – y – z = log₁₀ 12 – log₄ 2 × log₁₀ 9 – log₁₀ 0.4

= log₁₀ (3×4) – log₄ 4^1/2 × log₁₀ 3² – log₁₀ 4/10

= log₁₀ 3 + log₁₀ 4 – ½ log₄ 4 × 2 log₁₀ 3 – (log₁₀ 4 – log₁₀ 10)

= log₁₀ 3 + log₁₀ 4 – ½ × 1 × 2 log₁₀ 3 – log₁₀ 4 + 1

= log₁₀ 3 + log₁₀ 4 – log₁₀ 3 – log₁₀ 4 + 1

= 1

(ii) 7^x-y-z

7^x-y-z = 7¹

= 7

Question 10. If log V + log 3 = log π + log 4 + 3 log r, find V in terms of other quantities.

Answer:

Given:

log V + log 3 = log π + log 4 + 3 log r

Let us simplify the given expression to find V,

log (V × 3) = log (π × 4 × r³)

log 3V = log 4πr³

3V = 4πr³

V = 4πr³/3

Question 11. Given 3 (log 5 – log 3) – (log 5 – 2 log 6) = 2 – log n, find n.

Answer:

Given:

3 (log 5 – log 3) – (log 5 – 2 log 6) = 2 – log n

Let us simplify the given expression to find n,

3 log 5 – 3 log 3 – log 5 + 2 log 6 = 2 – log n

2 log 5 – 3 log 3 + 2 log 6 = 2 (1) – log n

log 5² – log 3³ + log 6² = 2 log 10 – log n [Since, 1 = log 10]

log 25 – log 27 + log 36 – log 10² = – log n

log n = – log 25 + log 27 – log 36 + log 100

= (log 100 + log 27) – (log 25 + log 36)

= log (100×27) – log (25×36)

= log (100×27)/(25×36)

log n = log 3

n = 3

Question 12. Given that log₁₀ y + 2 log₁₀ x = 2, express y in terms of x.

Answer:

Given:

log₁₀ y + 2 log₁₀ x = 2

Let us simplify the given expression,

log₁₀ y + log₁₀ x² = 2(1)

log₁₀ y + log₁₀ x² = 2 log₁₀ 10

log₁₀ (y×x²) = log₁₀ 10²

yx² = 100

y = 100/x²

Question 13. Express log₁₀ 2 + 1 in the form log₁₀x.

Answer:

Given:

log₁₀ 2 + 1

Let us simplify the given expression,

log₁₀ 2 + 1 = log₁₀ 2 + log₁₀ 10 [Since, 1 = log₁₀ 10 ]

= log₁₀ (2×10)

= log₁₀ 20

Question 14. If a² = log₁₀ x, b² = log₁₀ y and a²/2 – b²/3 = log₁₀ z. Express z in terms of x and y.

Answer:

Given:

a² = log₁₀ x

b² = log₁₀ y

a²/2 – b²/3 = log₁₀ z

Let us substitute the given values in the expression, we get

log₁₀ x/2 – log₁₀ y/3 = log₁₀ z

log₁₀ x^1/2 – log₁₀ y^1/3 = log₁₀ z

log₁₀ √x – log₁₀ ∛y = log₁₀ z

log₁₀ √x/∛y = log₁₀ z

√x/∛y = z

z = √x/∛y

Question 15. Given that log m = x + y and log n = x – y, express the value of log m²n in terms of x and y.

Answer:

Given:

log m = x + y

log n = x – y

log m²n

Let us simplify the given expression,

log m²n = log m² + log n

= 2 log m + log n

By substituting the given values, we get

= 2 (x + y) + (x – y)

= 2x + 2y + x – y

= 3x + y

Question 16. Given that log x = m + n and log y = m – n, express the value of log (10x/y²) in terms of m and n.

Answer:

Given:

log x = m + n

log y = m – n

log (10x/y²)

Let us simplify the given expression,

log (10x/y²) = log 10x – log y²

= log 10 + log x – 2 log y

= 1 + log x – 2 log y

= 1 + (m + n) – 2(m – n)

= 1 + m + n – 2m + 2n

= 1 – m + 3n

Question 17. If log x/2 = log y/3, find the value of y⁴/x⁶.

Answer:

Given:

log x/2 = log y/3

Let us simplify the given expression,

By cross multiplying, we get

3 log x = 2 log y

log x³ = log y²

so, x³ = y²

now square on both sides, we get

(x³)² = (y²)²

x⁶ = y⁴

y⁴/x⁶ = 1

Question 18. Solve for x:

(i) log x + log 5 = 2 log 3

(ii) log₃ x – log₃ 2 = 1

(iii) x = log 125/log 25

(iv) (log 8/log 2) × (log 3/log√3) = 2 log x

Answer:

(i) log x + log 5 = 2 log 3

Let us solve for x,

Log x = 2 log 3 – log 5

= log 3² – log 5

= log 9 – log 5

= log (9/5)

∴ x = 9/5

(ii) log₃ x – log₃ 2 = 1

Let us solve for x,

log₃ x = 1 + log₃ 2

= log₃ 3 + log₃ 2 [Since, 1 can be written as log₃ 3 = 1]

= log₃ (3×2)

= log₃ 6

∴ x = 6

(iii) x = log 125/log 25

x = log 5³/log5²

= 3 log 5/ 2 log 5

= 3/2 [Since, log 5/log 5 = 1]

∴ x = 3/2

(iv) (log 8/log 2) × (log 3/log√3) = 2 log x

(log 2³/log 2) × (log 3/log3^1/2) = 2 log x

(3log 2/log 2) × (log 3/½ log 3) = 2 log x

3 × 1/(½) = 2 log x

3 × 2 = 2 log x

6 = 2 log x

log x = 6/2

log x = 3

x = (10)³

= 1000

∴ x = 1000

Question 19. Given 2 log₁₀ x + 1= log₁₀ 250, find

(i) x

(ii) log₁₀2x

Answer:

Given:

2 log₁₀ x + 1= log₁₀ 250

(i) let us simplify the above expression,

log₁₀ x² + log₁₀ 10 = log₁₀ 250 [Since, 1 can be written as log₁₀ 10]

log₁₀ (x² × 10) = log₁₀ 250

(x² × 10) = 250

x² = 250/10

x² = 25

x = √25

= 5

∴ x = 5

(ii) log₁₀ 2x

We know that, x = 5

So, log₁₀ 2x = log₁₀ 2×5

= log₁₀ 10

= 1

Question 20. If log x/log 5 = log y²/log 2 = log 9/log (1/3), find x and y.

Answer:

Given:

log x/log 5 = log y²/log 2 = log 9/log (1/3)

let us consider,

log x/log 5 = log 9/log (1/3)

log x = (log 9×log 5)/log (1/3)

= (log 3² × log 5) / (log 1 – log 3)

= (2 log 3 × log 5) / (-log 3) [log 1 = 0]

= -2 × log 5

= log 5^-2

x = 5^-2

= 1/5²

= 1/25

Now,

log y²/log 2 = log 9/log (1/3)

log y² = (log 9×log2)/log (1/3)

= (log 3² × log 2) / (log 1 – log 3)

= (2 log 3 × log 2) / (-log 3) [log 1 = 0]

= -2 × log 2

= log 2^-2

y² = 2^-2

= 1/2²

= ¼

= √(1/4)

= ½

Question 21. Prove the following:

(i) 3^{log 4} = 4^{log 3}
(ii) 27^{log 2} = 8^{log 3}

Answer:

(i) 3^{log 4} = 4^{log 3}

Let us take log on both sides,

If log 3^{log 4} = log 4^{log 3}

log 4 . log 3 = log 3 . log 4

log 2² . log 3 = log 3 . log 2²

2 log 2 . log 3 = log 3 . 2 log 2

Which is true.

Hence proved.

(ii) 27^{log 2} = 8^{log 3}

Let us take log on both sides,

If log 27^{log 2} = log 8^{log 3}

log 2 . log 27 = log 3 . log 8

log 2 . log 3³ = log 3 . log 2³

log 2 . 3 log 3 = log 3 . 3 log 2

3 log2 . log 3 = 3 log2 . log 3

Which is true.

Hence proved.

Question 22. Solve the following equations:

(i) log (2x + 3) = log 7

(ii) log (x +1) + log (x – 1) = log 24

(iii) log (10x + 5) – log (x – 4) = 2

(iv) log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + 1

(v) log (4y – 3) = log (2y + 1) – log 3

(vi) log₁₀ (x + 2) + log₁₀ (x – 2) = log₁₀3 + 3 log₁₀ 4

(vii) log (3x + 2) + log (3x – 2) = 5 log 2

Answer:

(i) log (2x + 3) = log 7

Let us simplify the expression,

2x + 3 = 7

2x = 7 – 3

2x = 4

x = 4/2

= 2

(ii) log (x +1) + log (x – 1) = log 24

Let us simplify the expression,

log [(x +1) (x – 1)] = log 24

log (x² – 1) = log 24

(x² – 1) = 24

x² = 24 + 1

= 25

x = √25

= 5

(iii) log (10x + 5) – log (x – 4) = 2

Let us simplify the expression,

log (10x + 5) / (x – 4) = 2 log 10

log (10x + 5) / (x – 4) = log 10²

(10x + 5) / (x – 4) = 100

10x + 5 = 100 (x – 4)

10x + 5 = 100x – 400

5 + 400 = 100x – 10x

90x = 405

x = 405/90

= 81/18

= 9/2

= 4.5

(iv) log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + 1

Let us simplify the expression,

log₁₀ [5× (5x + 1)] = log₁₀ (x + 5) + log₁₀ 10

log₁₀ [5× (5x + 1)] = log₁₀ [(x + 5) × 10] [5× (5x + 1)] = [(x + 5) × 10]

25x + 5 = 10x + 50

25x – 10x = 50 – 5

15x = 45

x = 45/15

= 3

(v) log (4y – 3) = log (2y + 1) – log 3

Let us simplify the expression,

log (4y – 3) = log (2y + 1) / 3

(4y – 3) = (2y + 1) / 3

By cross multiplying, we get

3(4y – 3) = 2y + 1

12y – 9 = 2y + 1

12y – 2y = 9 + 1

10y = 10

y = 10/10

= 1

(vi) log₁₀ (x + 2) + log₁₀ (x – 2) = log₁₀3 + 3 log₁₀ 4

Let us simplify the expression,

log₁₀ [(x + 2) × (x – 2)] = log₁₀ 3 + log₁₀ 4³

log₁₀ [(x + 2) × (x – 2)] = log₁₀ (3×4³)

[(x + 2) × (x – 2)] = (3×4³)

(x² – 4) = (3×4×4×4)

(x² – 4) = 192

x² = 192 + 4

= 196

x = √196

= 14

(vii) log (3x + 2) + log (3x – 2) = 5 log 2

Let us simplify the expression,

log (3x + 2) + log (3x – 2) = log 2⁵

log [(3x + 2) × (3x – 2)] = log 32

log (9x² – 4) = log 32

(9x² – 4) = 32

9x² = 32 + 4

9x² = 36

x² = 36/9

x² = 4

x = √4

= 2

Question 23. Solve for x:

log₃ (x + 1) – 1 = 3 + log₃ (x – 1)
Answer:

Given:

log₃ (x + 1) – 1 = 3 + log₃ (x – 1)

Let us simplify the expression,

log₃ (x + 1) – log₃ (x – 1) = 3 + 1

log₃ (x + 1) / (x – 1) = 4 log₃ 3 [Since, log₃ 3 = 1]

log₃ (x + 1) / (x – 1) = log₃ 3⁴

(x + 1) / (x – 1) = 3⁴

By cross multiplying, we get

(x + 1) = 81 (x – 1)

x + 1 = 81x – 81

81x – x = 1 + 81

80x = 82

x = 82/80

= 41/40

= 1 1/40

Question 24. Solve for x:

5^{log x} + 3^{log x} = 3^{log x+1} – 5^{log x – 1}

Answer:

Given:

5^{log x} + 3^{log x} = 3^{log x+1} – 5^{log x – 1}

Let us simplify the expression,

5^{log x} + 3^{log x} = 3^{log x} . 3¹ – 5^{log x} . 5^-1

5^{log x} + 3^{log x} = 3.3^{log x} – 1/5 . 5^{log x}

5^{log x} + 1/5 . 5^{log x} = 3.3^{log x} – 3^{log x}

(1 + 1/5) 5^{log x} = (3 – 1) 3^{log x}

(6/5) 5^{log x} = 2(3^{log x})

5^{log x} / 3^{log x} = (2×5)/6

(5/3)^{log x} = 10/6

(5/3)^{log x} = 5/3

(5/3)^{log x} = (5/3)¹

So, by comparing the powers

log x = 1

log x = log 10

x = 10

Question 25. If log (x-y)/2 = ½ (log x + log y), prove that x² + y² = 6xy

Answer:

Given:

log (x-y)/2 = ½ (log x + log y)

Let us simplify,

log (x-y)/2 = ½ (log x×y)

log (x-y)/2 = ½ log xy

log (x-y)/2 = log (xy)^1/2

(x-y)/2 = (xy)^1/2

By squaring on both sides, we get

[(x-y)/2]² = [(xy)^1/2]²

(x – y)²/4 = xy

By cross multiplying, we get

(x – y)² = 4xy

x² + y² – 2xy = 4xy

x² + y² = 4xy + 2xy

x² + y² = 6xy

Hence proved.

Question 26. If x² + y² = 23xy, Prove that log (x + y)/5 = ½ (log x + log y)

Answer:

Given:

x² + y² = 23xy

So, the above equation can be written as

x² + y² = 25xy – 2xy

x² + y² + 2xy = 25xy

(x + y)² = 25xy

(x + y)² / 25 = xy

Now by taking log on both sides, we get

log [(x + y)² / 25] = log xy

log [(x + y)/5]² = log xy

2 log (x+y)/5 = log x + log y

log (x+y)/5 = ½ log x + log y

Hence proved.

Question 27. If p = log₁₀ 20 and q = log₁₀ 25, find the value of x if 2 log₁₀ (x + 1) = 2p – q

Answer:

Given:

p = log₁₀ 20

q = log₁₀ 25

Then,

2 log₁₀ (x + 1) = 2p – q

Now substitute the values of p and q, we get

2 log₁₀ (x + 1) = 2 log₁₀ 20 – log₁₀ 25

= 2 log₁₀ 20 – log₁₀ 5²

= 2 log₁₀ 20 – 2 log₁₀ 5

2 log₁₀ (x + 1) = 2 (log₁₀ 20 – log₁₀ 5)

log₁₀ (x + 1) = (log₁₀ 20 – log₁₀ 5)

= log₁₀ (20/5)

log₁₀ (x + 1) = log₁₀ 4

(x + 1) = 4

x = 4 – 1

= 3

Question 28. Show that:

(i) 1/log₂ 42 + 1/log₃ 42 + 1/log₇ 42 = 1

(ii) 1/log₈ 36 + 1/log₉ 36 + 1/log₁₈ 36 = 2

Answer:

(i) 1/log₂ 42 + 1/log₃ 42 + 1/log₇ 42 = 1

Let us consider LHS:

1/log₂ 42 + 1/log₃ 42 + 1/log₇ 42

By using the formula, log_n m = log_m / log_n

1/log₂ 42 + 1/log₃ 42 + 1/log₇ 42 = 1/(log 42/log₂) + 1/(log 42/log₃) + 1/(log 42/log₇)

= log₂/log 42 + log₃/log 42 + log₇/log 42

= (log₂ + log₃ + log₇)/log 42

= (log 2×3×7)/log 42

= log 42 / log 42

= log 42/log 42

= 1

= RHS

(ii) 1/log₈ 36 + 1/log₉ 36 + 1/log₁₈ 36 = 2

Let us consider LHS:

1/log₈ 36 + 1/log₉ 36 + 1/log₁₈ 36

By using the formula, log_n m = log_m / log_n

1/log₈ 36 + 1/log₉ 36 + 1/log₁₈ 36 = 1/(log 36/log₈) + 1/(log 36/log₉) + 1/(log 36/log₁₈)

= log₈/log 36 + log₉/log 36 + log₁₈/log 36

= (log₈ + log₉ + log₁₈)/log 36

= (log 8×9×18)/log 36

= log 36²/log 36

= 2 log 36/log 36

= 2

= RHS

Question 29. Prove the following identities:

(i) 1/log_a abc + 1/log_b abc + 1/log_c abc = 1

(ii) log_b a. log_c b. log_d c = log_d a

Answer:

(i) 1/log_a abc + 1/log_b abc + 1/log_c abc = 1

Let us consider LHS:

1/log_a abc + 1/log_b abc + 1/log_c abc

By using the formula, log_n m = log_m / log_n

1/log_a abc + 1/log_b abc + 1/log_c abc = 1/(log abc/log_a) + 1/(log abc/log_b) + 1/(log abc/log_c)

= log_a/ log abc + log_b/ log abc + log_c/ log abc

= (log_a + log_b + log_c)/ log abc

= (log a×b×c)/ log abc

= log abc/log abc

= 1

= RHS

(ii) log_b a. log_c b. log_d c = log_d a

Let us consider LHS:

log_b a. log_c b. log_d c = (log a/ log b) × (log b/ log c) × (log c/ log d)

= log a/log d

= log_d a

= RHS

Question 30. Given that log_a x = 1/ α, log_b x = 1/β, log_c x = 1/γ, find log_abc x.

Answer:

It is given that:

log_a x = 1/ α, log_b x = 1/β, log_c x = 1/γ

So,

log_a x = 1/ α => log x/log_a = 1/ α => log_a = α log x

log_b x = 1/β => log x/log_b = 1/ β => log_b = β log x

log_c x = 1/γ => log x/log_c = 1/ γ => log_c = γ log x

Now,

log_abc x = log x/log abc

= log x/(log a + log b + log c)

= log x/(α log x + β log x + γ log x)

= log x/log x(α+ β+ γ)

= 1/(α+ β+ γ)

Question 31. Solve for x:

(i) log₃ x + log₉ x + log₈₁ x = 7/4

(ii) log₂ x + log₈ x + log₃₂ x = 23/15

Answer:

(i) log₃ x + log₉ x + log₈₁ x = 7/4

let us simplify the expression,

1/log_x 3 + 1/log_x 9 + 1/log_x 81 = 7/4

1/log_x 3¹ + 1/log_x 3² + 1/log_x 3⁴ = 7/4

1/log_x 3 + 1/2log_x 3 + 1/4log_x 3 = 7/4

1/log_x 3 [1 + ½ + ¼] = 7/4

1/log_x 3 [(4+2+1)/4] = 7/4

log₃ x [7/4] = 7/4

log₃ x = (7/4) × (4/7)

log₃ x = 1

log₃ x = log₃ 3 [Since, 1= log_a a]

On comparing, we get

x = 3

(ii) log₂ x + log₈ x + log₃₂ x = 23/15

let us simplify the expression,

1/log_x 2 + 1/log_x 8 + 1/log_x 32 = 23/15

1/log_x 2¹ + 1/log_x 2³ + 1/log_x 2⁵ = 23/15

1/log_x 2 + 1/3log_x 2 + 1/5log_x 2 = 23/15

1/log_x 2 [1 + 1/3 + 1/5] = 23/15

log₂ x [(15 + 5 + 3)/15] = 23/15

log₂ x [23/15] = 23/15

log₂ x = (23/15) × (15/23)

log₂ x = 1

log₂ x = log₂ 2 [Since, 1= log_a a]

On comparing, we get

x = 2

—  : End of ML Aggarwal Logarithms Exe-9.2 Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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