Class 9 Advanced Maths Chapter 2 Logarithms Ex 2.4 Solutions - #NCSOLVE 📚

Accurate and updated NCERT Class 9 Advanced Maths Solutions Chapter 2 Logarithms Ex 2.4 help students build a strong mathematical foundation.

Ex 2.4 Class 9 Advanced Maths Solutions

Advanced Maths Class 9 Exercise 2.4 Solutions

Exercise 2.4 Class 9 Advanced Maths Solutions

Question 1.
Solve for x.
(i) log₃ (2x – 5) = 2
(ii) log₇ (3x) + log₇ 2 = log₇ 24
(iii) log₅ (x + 3) – log 5 (x – 1) = 1
(iv) log₂ (x² – 7) = 3
Solution:
(i) Given, log₃ (2x – 5) = 2
⇒ 2x – 5 = 3² [∵ log_a x = n ⇒ x = aⁿ]
⇒ 2x – 5 = 9
⇒ 2x = 14
⇒ x = 7

(ii) Given, log₇ (3x) + log₇ 2 = log₇ 24 .
⇒ log₇ (3x . 2) = log₇ 24
[∵ log_a m + log_a n = log_a mn]
⇒ 6x = 24 [∵ log_a x = log_a y ⇒ x = y]
⇒ x = 4

(iii) Given, log₅ (x + 3) – log₅ (x – 1) = 1
⇒ log₅(\(\frac{x+3}{x-1}\)) = 1 [∵ log_a m + log_a n = log_a \(\frac{m}{n}\)]
\(\frac{x+3}{x-1}\) = 5¹ [∵ log_a x = n ⇒ x = aⁿ]
⇒ x + 3 = 5(x – 1)
⇒ x + 3 = 5x – 5
⇒ 4x = 8
⇒ x = 2

(iv) Given, log₂ (x² – 7) = 3
⇒ x² – 7 = 2³ [∵ log_a x = n ⇒ x = aⁿ]
⇒ x² – 7 = 8
⇒ x² = 15
⇒ x = ±\(\sqrt{15}\)

Question 2.
Solve for x.
(i) log₂ (x – 3) + log₂ (x + 1) = 5
(ii) 2 log₄ x = log₄ (5x – 4)
(iii) log₅ (x + 2) + log₅ (x – 2) = 1
(iv) log₁₀ (x – 2) + log₁₀ (x + 1) = 1
Solution:
(i) Given, log₂ (x – 3) + log₂ (x + 1) = 5
⇒ log₂ [(x – 3) (x + 1)] = 5 [∵ log_a m + log_a n = log_a mn]
⇒ (x – 3) (x + 1) = 2⁵ [∵ log_a x = n ⇒ x = aⁿ]
⇒ x² – 2x – 3 = 32
⇒ x² – 2x – 35 = 0
⇒ (x – 7)(x + 5) = 0
⇒ x = 7 or x = -5
But for x = -5, the value of log₂ (x – 3) is not defined. [∵ argument must be > 0]
∴ x = 7

(ii) Given, 2 log₄ x = log₄ (5x – 4)
⇒ log₄ x² = log₄(5x – 4) [∵ n log_a m = log_a mⁿ]
⇒ x² = 5x – 4 [∵ log_a x = log_a y ⇒ x = y]
⇒ x² – 5x + 4 = 0
⇒ (x – 4) (x – 1) = 0
⇒ x = 4, 1
For both x = 4 and x = 1, log₄ x is defined. [∵ x > 0]
∴ x = 1, 4

(iii) Given, log₅ (x + 2) + log₅ (x – 2) = 1
⇒ log_a [(x + 2) (x – 2)] = 1 [∵ log_a m + log_a n = log_a mn]
⇒ x² – 4 = 5¹ [∵ log_a x = n ⇒ x = aⁿ]
⇒ x² = 9 ⇒ x =± 3
But, for x = – 3, log₅ (x – 2) is not defined. [∵ argument must be > 0]
∴ x = 3

(iv) Given, log₁₀ (x – 2) + log₁₀ (x + 1) = 1
⇒ log₁₀ [(x – 2)(x + 1)] = 1
[∵ log_a m + log_a n = log_a mn]
⇒ (x – 2) (x + 1) = 10¹ [∵ log_a x = n ⇒ x = aⁿ]
⇒ x² – x – 2 = 10
⇒ x² – x – 12 = 0
⇒ (x – 4) (x + 3) = 0
⇒ x = 4 or x = -3
But, for x = – 3, log₁₀ (x – 2) is not defined.
[∵ argument must be > 0]
∴ x = 4

Question 3.
Solve for x.
(i) log_x (3x + 10) = 2, where x > 0 and x ≠ 1.
(ii) (log₃ x)² – 4 log₃ x + 3 = 0
(iii) (log₂ x)² + log₂ x³ = 10
(iv) x^{log10 x} = 1000x²
Solution:
(i) Given, log_x (3x + 10) = 2
⇒ 3x + 10 = x² [∵ log_a x = n ⇒ x = aⁿ]
⇒ x² – 3x – 10 = 0 ⇒ (x – 5) (x + 2) = 0
⇒ x = 5 or x = -2
∵ It is given that x > 0 and x ≠ 1.
∴ x = 5

(ii) Given, (log₃ x)² – 4 log₃ x + 3 = 0
Let y = log₃ x, the given equation reduces to
y² – 4y + 3 = 0 ⇒ (y – 3)(y – 1) = 0
⇒ y = 3 or y = 1
∴ log₃ x = 3 or log 3 x = 1
⇒ x = 3³ or x = 3¹ [∵ log_a x = n ⇒ x = aⁿ]
⇒ x = 27, 3

(iii) Given, (log₂ x)² + log₂ x³ = 10
⇒ (log₂ x)² + 3 log₂ x = 10 [∵ log_a mⁿ = n log_a m]
Let y = log₂ x, the given equation reduces to
y² + 3y – 10 = 0 ⇒ (y + 5) (y – 2) = 0
⇒ y = -5 or y = 2
log₂x = -5 or log₂ x = 2
⇒ x = 2^-5 or x = 2² [∵ log_a x = n
⇒ x = aⁿ]
⇒ x = \(\frac{1}{32}\) , 4

(iv) Given x^{log10 x} = 1000x²
On taking log₁₀ on both sides, we get
⇒ log₁₀(x^{log10 x}) = log₁₀ (1000x²) [∵ log_a mn = log_a m + log_a n]
⇒ (log₁₀ x)² = log₁₀ 10³ + log₁₀ x²
⇒ (log₁₀x)² = 3 log₁₀ 10 + 2 log₁₀ x [∵ log_a mⁿ = n log_a m]
⇒ (log₁₀ x)² = 3 + 2 log₁₀ x
⇒ (log₁₀ x)² – 2 log₁₀ x – 3 = 0
Let y = log₁₀ x, we get y² – 2y – 3 = 0
⇒ (y – 3) 0 + 1) = 0
⇒ y = 3 or y = -1
⇒ log₁₀ = 3 or log₁₀ x = -1
⇒ x = 10³ or x = 10^-1 [∵ log_a x = n ⇒ x = aⁿ]
⇒ x = 1000, 0.1

Question 4
Solve for x.
(i) log₃ (x² – 1) = log₃ (2x – 1)
(ii) log_x 5 – log 2 = log_x √x
(iii) log₂ x + \(\frac{1}{\log _x 2}\) = 4
(iv) log₃ (3 + x) + log₃ (8 – x) – log₃ (9x – 8) = 2 – log₃ 9
(v) log₁₀ [log₂(log₃ 9)] = 5x
Solution:

(i) Given, log₃ (x² – 1) = log₃ (2x – 1)
⇒ x² – 1 = 2x – 1 [∵ log_a x = log_a y ⇒ x = y]
⇒ x² – 2x = 0 ⇒ x(x – 2) = 0
⇒ x = 0 or x = 2
But, for x = 0, log₃ (2x – 1) = log₃ (-1), which is not
defined.
∴ x = 2

(ii) Given, log_x 5 – log_x 2 = log_x √x
⇒ log_x (\(\frac{5}{2}\)) = log_x √x [∵ log_a m – log_a n = log_a \(\frac{m}{n}\)]
⇒ (\(\frac{5}{2}\)) = √x [∵ log_a x = log_a y ⇒ x = y]
On squaring both sides, we get
⇒ x = (\(\frac{5}{2}\))²
⇒ x = \(\frac{25}{4}\) or 6.25

(iii) Given log₂ x + \(\frac{1}{\log _x 2}\) = 4
⇒ log₂ x + log₂ x = 4 [∵ \(\frac{1}{\log _a b}\) = log_b a]
⇒ 2 log₂ x = 4
⇒ log₂ x = 2
⇒ x = 2² [∵ log_a x = n ⇒ x = aⁿ]
⇒ x = 4

(iv) Given, log₃ (3 + x) + log₃ (8 – x) – log₃ (9x – 8) = 2 – log₃ 9
⇒ log₃ [latex]\frac{(3+x)(8-x)}{9 x-8}[/latex] = 2 – 2 [∵ log₃ 9 = 2]
⇒ log₃ [latex]\frac{24+5 x-x^2}{9 x-8}[/latex] = 0
⇒ \(\frac{24+5 x-x^2}{9 x-8}\) = 3⁰ [∵ log_a x = n ⇒ x = aⁿ
⇒ 24 + 5x – x² = 1(9x – 8)
⇒ x² + 4x – 32 = 0
⇒ (x + 8) (x – 4) = 0
⇒ x = -8 or x = 4
Checking original arguments, for x = -8,(3 + x) is negative.
∴ x = 4

(v) Given, log₁₀ [log₂(log₃ 9)] = 5x
⇒ log₁₀[log₂(2)] = 5x [∵ log₃ 9 = 2]
⇒ log₁₀(1) = 5x [∵ log_a a = 1]
⇒ 0 = 5x [∵ log_a 1 = 0]
⇒ x = 0

Question 5.
If x = log\(\frac{1}{2}\) + log\(\frac{2}{3}\) + log\(\frac{3}{4}\) +… + log\(\frac{99}{100}\), where all logs are to the base 10 then evaluate (x + 1) (x + 2) (x + 3) … (x + 99).
Solution:
Given,
x = log₁₀ \(\frac{1}{2}\) + log₁₀ \(\frac{2}{3}\) + log₁₀ \(\frac{3}{4}\) +… + log₁₀ \(\frac{99}{100}\),
⇒ x = log₁₀ (\(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{3}{4}\), …..\(\frac{99}{100}\)) [∵ logm + logn = logmn]
⇒ x = log₁₀(\(\frac{1}{100}\))
⇒ x = log₁₀(10^-2)
⇒ x = -2 log₁₀ 10 [∵ log mⁿ = nlogm]
⇒ x = -2 [∵ log_a a = 1]
Now, let E = (x + 1) (x + 2) (x + 3) … (x + 99)
On substituting x = -2 in the expression, we get
E = (-2 + 1) (-2 + 2) (-2 + 3) … (-2 + 99)
⇒ E = (-1) (0) (1) … (97)
⇒ E = 0 [∵ multiplication by zero results in zero]
∴ (x + 1) (x + 2) (x + 3) … (x + 99) = 0

The post Class 9 Advanced Maths Chapter 2 Logarithms Ex 2.4 Solutions appeared first on Learn CBSE.

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