Chapter 9 Logarithms ML Aggarwal ICSE Solutions for Class 9 Maths

ML Aggarwal Solutions for Chapter 9 Logarithms Class 9 Maths ICSE

Here, we are providing the solutions for Chapter 9 Logarithms from ML Aggarwal Textbook for Class 9 ICSE Mathematics. Solutions of the ninth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 9 Chapter 9 Logarithms ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on converting to logarithmic form from exponential form, converting to exponential form from logarithmic form and solving the equations of log for finding the value.

Exercise 9.1

1. Convert the following to logarithmic form:
(i) 5² = 25
(ii) a⁵ = 64
(iii) 7^x = 100
(iv) 9^o = 1
(v) 6¹ = 6
(vi) 3^-2 = 1/9
(vii) 10^-2 = 0.01
(viii) (81)^3/4 = 27

Solution

(i) 5² = 25

Let us apply log, we get

Log₅ 25 = 2

(ii) a⁵ = 64

Let us apply log, we get

Log_a 64 = 5

(iii) 7^x = 100

Let us apply log, we get

Log₇ 100 = x

(iv) 9^o = 1

Let us apply log, we get

Log₉ 1 = 0

(v) 6¹ = 6

Let us apply log, we get

Log₆ 6 = 1

(vi) 3^-2 = 1/9

Let us apply log, we get

Log₃ 1/9 = -2

(vii) 10^-2 = 0.01

Let us apply log, we get

Log₁₀ 0.01 = -2

(viii) (81)^3/4 = 27

Let us apply log, we get

Log₈₁ 27 = ¾

2. Convert the following into exponential form:

(i) log₂ 32 = 5

(ii) log₃ 81=4

(iii) log₃ 1/3 = -1

(iv) log₃ 4 = 2/3

(v) log₈ 32 = 5/3

(vi) log₁₀ (0.001) = -3

(vii) log₂ 0.25 = -2

(viii) log_a (1/a) = -1

Solution

(i) log₂ 32 = 5

The exponential form of the expression is

2⁵ = 32

(ii) log₃ 81=4

The exponential form of the expression is

3⁴ = 81

(iii) log₃ 1/3 = -1

The exponential form of the expression is

3^-1 = 1/3

(iv) log₃ 4 = 2/3

The exponential form of the expression is

(8)^2/3 = 4

(v) log₈ 32 = 5/3

The exponential form of the expression is

(8)^5/3 = 32

(vi) log₁₀ (0.001) = -3

The exponential form of the expression is

10^-3 = 0.001

(vii) log₂ 0.25 = -2

The exponential form of the expression is

2^-2 = 0.25

(viii) log_a (1/a) = -1

The exponential form of the expression is

a^-1 = 1/a

3. By converting to exponential form, find the values of:

(i) log₂ 16

(ii) log₅ 125

(iii) log₄ 8

(iv) log₉ 27

(v) log₁₀ (.01)

(vi) log₇ 1/7

(vii) log.₅ 256

(viii) log₂ 0.25

Solution

(i) log₂ 16

Let us consider log₂ 16 = x

So,

2^x = 16

= 2 × 2 × 2 × 2

2^x = 2⁴

By comparing the powers,

x = 4

(ii) log₅ 125

Let us consider log₅ 125 = x

So,

(5)^x = 125

= 5 × 5 × 5

5^x = 5³

By comparing the powers,

x = 3

(iii) log₄ 8

Let us consider log₄ 8 = x

So,

(4)^x = 8

(2 × 2)^x = 2 × 2 × 2

2^2x = 2³

By comparing the powers,

2x = 3

x = 3/2

(iv) log₉ 27

Let us consider log₉ 27= x

So,

(9)^x = 27

(3 × 3)^x = 3 × 3 × 3

3^2x = 3³

By comparing the powers,

2x = 3

x = 3/2

(v) log₁₀ (.01)

Let us consider log₁₀ (.01) = x

So,

(10)^x = 1/100

= 1/10 × 1/10

10^x = 1/(10)²

10^x = 10^-2

By comparing the powers,

x = -2

(vi) log₇ 1/7

Let us consider log₇ 1/7 = x

So,

(7)^x = 1/7

7^x = 7^-1

By comparing the powers,

x = -1

(vii) log.₅ 256

Let us consider log.₅ 256 = x

So,

(.5)^x = 256

(5/10)^x = 2×2×2×2×2×2×2×2

(1/2)^x = 2⁸

(2)^-x = 2⁸

By comparing the powers,

-x = 8

x = -8

(viii) log₂ 0.25

Let us consider log₂ 0.25 = x

So,

(2)^x = 0.25

= 25/100

2^x = 1/4

2^x = (2)^-2

By comparing the powers,

x = -2

4. Solve the following equations for x:

(i) log₃ x = 2

(ii) log_x 25 = 2

(iii) log₁₀ x = -2

(iv) log₄ x = ½

(v) log_x 11 = 2.5

(vi) log_x ¼ = -1

(vii) log₈₁ x = 3/2

(viii) log₉ x = 2.5

(ix) log₄ x = -1.5

Solution

(i) log₃ x = 2

Let us simplify the expression,

(3)² = x

x = 9

(ii) log_x 25 = 2

Let us simplify the expression,

(x)² = 25

= 5×5

x² = 5²

Since the powers are same,

So,

x = 5

(iii) log₁₀ x = -2

Let us simplify the expression,

(10)^-2 = x

x = 1/(10)²

= 1/100

x = 0.01

(iv) log₄ x = ½

Let us simplify the expression,

(4)^1/2 = x

x = (2 × 2)^1/2

= (2)²×^1/2

x = 2

(v) log_x 11 = 2.5

Let us simplify the expression,

(x)¹ = 11

x = 11

(vi) log_x ¼ = -1

Let us simplify the expression,

(x)^-1 = ¼

x^-1 = 4^-1

Since the powers are same,

So,

x = 4

(vii) log₈₁ x = 3/2

Let us simplify the expression,

(81)^3/2 = x

x = 81^3/2

= (3⁴)^3/2

= 3⁴×^3/2

= 3⁶

= 3×3×3×3×3×3

= 729

x = 729

(viii) log₉ x = 2.5

log₉ x = 5/2

Let us simplify the expression,

(9)^5/2 = x

x = (3²)^5/2

= 3²×^5/2

= 3⁵

= 3×3×3×3×3

= 234

x = 234

(ix) log₄ x = -1.5

log₄ x = -3/2

Let us simplify the expression,

(4)^-3/2 = x

x = (2²)^-3/2

= 2²×^-3/2

= 2^-3

= 1/2³

= 1/(2×2×2)

= 1/8

x = 1/8

5. Given log₁₀ a = b, express 10^2b-3 in terms of a.

Solution

Given:

log₁₀ a = b

(10)^b = a

Now,

10^2b-3 = (10)^2b / (10)³

= (10^b)²/(10×10×10)

Substitute the value of (10)^b = a, we get

= a²/1000

6. Given log₁₀ x= a, log₁₀ y = b and log₁₀ z =c,

(i) write down 10^2a-3 in terms of x.

(ii) write down 10^3b-1 in terms of y.

(iii) if log₁₀ P = 2a + b/2 – 3c, express P in terms of x, y and z.

Solution

Given:

log₁₀ x= a

• (10)^a = x

log₁₀ y = b

• (10)^b = y

log₁₀ z =c

• (10)^c = z

(i) Write down 10^2a-3 in terms of x.

10^2a-3 = (10)^2a/(10)³

= (10^a)²/(10×10×10)

Substitute the value of (10)^a = x, we get

= x²/1000

(ii) Write down 10^3b-1 in terms of y.

10^3b-1 = (10)^3b/(10)¹

= (10^b)³/(10)

Substitute the value of (10)^b = y, we get

= y³/10

(iii) If log₁₀ P = 2a + b/2 – 3c, express P in terms of x, y and z.

we know that,

(10)^a = x

(10)^b = y

(10)^c = z

By substituting the values

log₁₀ P = 2a + b/2 – 3c

= 2 log₁₀ x + ½ log₁₀ y – 3 log₁₀ z

= log₁₀ x² + log₁₀ y^1/2 – log₁₀ z³

= log₁₀ (x² + y^1/2) – log₁₀ z³

= log₁₀ [(x²√y)/z³]

P = (x²√y)/z³ 

7. If log₁₀x = a and log₁₀y = b, find the value of xy.

Solution

Given:

log₁₀x = a

⇒ (10)^a = x

⇒ log₁₀y = b

⇒ (10)^b = y

Then,

xy = (10)^a × (10)^b

= (10)^a+b

8. Given log₁₀ a = m and log₁₀ b = n, express a³/b² in terms of m and n.

Solution

Given:

log₁₀ a = m

⇒ (10)^m = a

⇒ log₁₀ b = n

⇒ (10)ⁿ = b

So,

a³/b² = (10^m)³/(10ⁿ)²

= 10^3m/10²ⁿ

= 10^3m–2n

9. Given log₁₀a= 2a and log₁₀y = –b/2

(i) write 10^a in terms of x.

(ii) write 10^2b+1 in terms of y.

(iii) if log₁₀P= 3a – 2b, express P in terms of x and y.

Solution

Given:

log₁₀a= 2a

⇒ (10)^2a = a

⇒ log₁₀y = –b/2

⇒ (10)^-b/2 = y

(i) Write 10^a in terms of x.

10^a = (10^2a)^1/2

= (x)^1/2

= √x

(ii) Write 10^2b+1 in terms of y.

10^2b+1 = 10^2b ×10¹

= 10^4(b/2) × 10¹

= (10^b/2)⁴ ×10¹

= y⁴ × 10¹

= 10y⁴

(iii) If log₁₀P= 3a – 2b, express P in terms of x and y.

log₁₀P= 3a – 2b

Substitute the values,

log₁₀P= 3/2 (2a) – 4(b/2)

= 3/2 (log₁₀ x) – 4 (log₁₀ y)

= (log₁₀ x)^3/2 – (log₁₀ y)⁴

= log₁₀ [(x^3/2)/y⁴]

P = (x^3/2)/y⁴

10. If log₂ y = x and log₃ z = x, find 72^x in terms of y and z.

Solution

Given:

log₂ y = x

⇒ 2^x = y

⇒ log₃ z = x

⇒ 3^x = z

So,

72^x = (2×2×2×3×3)^x

= (2³ × 3²)^x

= 2^3x × 3^2x

= (2^x)³ × (3^x)²

= y³ × z²

= y³z²

11. If log₂ x = a and log₅y = a, write 100^2a-1 in terms of x and y.

Solution

Given:

log₂ x = a

⇒ 2^a = x

⇒ log₅y = a

⇒ 5^a = y

So,

100^2a-1 = (2×2×5×5)^2a-1

= (2² × 5²)^2a-1

= 2^4a-2 × 5^a-2

= (2^4a)/2² × (5^4a)/5²

= [(2^a)⁴ × (5^a)⁴]/(4×25)

= (x⁴y⁴)/100

Exercise 9.2

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

Solution

(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

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

Solution

(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

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

Solution

(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

4. Prove the following:

(i) log₁₀ 4 ÷ log₁₀ 2 = log₃ 9

(ii) log₁₀ 25 + log₁₀ 4 = log₅ 25

Solution

(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.

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

Solution

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

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

(i) x

(ii) log₁₀ ⁵√x²

(iii) log₁₀ 5x

Solution

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

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

Solution

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

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

Solution

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

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

Solution

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 – (log₁₀4 – log₁₀ 10)

= log₁₀ 3 + log₁₀ 4 – ½ × 1 × 2 log₁₀ 3 – log₁₀4 + 1

= log₁₀ 3 + log₁₀ 4 – log₁₀ 3 – log₁₀4 + 1

= 1

(ii) 7^x-y-z

7^x-y-z = 7¹

= 7

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

Solution

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

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

Solution

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

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

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

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

Solution

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

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

Solution

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

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

Solution

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

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.

Solution

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

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

Solution

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

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

Solution

(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

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

(i) x

(ii) log₁₀2x

Solution

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

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

Solution

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

21. Prove the following:

(i) 3^{log 4} = 4^{log 3}

(ii) 27^{log 2} = 8^{log 3}

Solution

(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.

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

Solution

(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) log₁₀5 + 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)] = log₁₀3 + log₁₀4³

⇒ 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

23. Solve for x:

log₃ (x + 1) – 1 = 3 + log₃ (x – 1)

Solution

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 log₃3 [Since, log₃ 3 = 1]

⇒ log₃(x + 1)/(x – 1) = log₃3⁴

⇒ (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

24. Solve for x:

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

Solution

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

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

Solution

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.

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

Solution

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.

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

Solution

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 log₁₀20 – 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

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

Solution

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

Let us consider LHS:

1/log₂42 + 1/log₃42 + 1/log₇42

By using the formula, log_nm = log_m/log_n

1/log₂42 + 1/log₃42 + 1/log₇42

= 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

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

Solution

(i) 1/log_aabc + 1/log_babc + 1/log_cabc = 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

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

Solution

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)

31. Solve for x:

(i) log₃ x + log₉ x + log₈₁ x = 7/4

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

Solution

(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