Test: Logarithm- 1 - CAT MCQ

• Since log_a a = 1, so log₁₀ 10 = 1.
• log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3 
  ∴ log (2 + 3) ≠ log (2 x 3)
• Since loga 1 = 0. so logio 1 = 0.
• log (1 + 2 + 3) = log 6 
   log (1 x 2 x 3) = log 1 + log 2 + log 3=log (6). 

So. option (b) is incorrect.

Given, log 27 = 1.431
⇒ log (3³) = 1.431
⇒ 3 log 3 = 1.431
⇒ log 3 = 0.477
∴ log 9 = log (3²) = 2 log 3
⇒ (2 x 0.477) = 0.954

⇒ - log₁₀ (7 x 10)
⇒ - (log₁₀ 7 + log₁₀ 10)
⇒ - (a + 1)

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

⇒ log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + log₁₀ 10

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

⇒ 5(5x + 1) = 10(x + 5)

⇒ 5x + 1 = 2x + 10

⇒ 3x = 9

⇒ x = 3.

Given expression = 1/log₆₀ 3 + 1/log₆₀ 4 + 1/log₆₀ 5
= log₆₀ (3 x 4 x 5)
= log₆₀ 60
= 1.

Given, log_x (x² + 12) = 4
⇒ x² + 12 = x⁴ 
⇒ a + 12 = a²     [Take x² = a]
⇒ a² – a – 12 = 0
⇒ (a – 4)(a + 3) = 0
⇒ a = -3 or 4.        [a = x² cannot be negative hence -3 is rejected]
⇒ x² = 4 
⇒ x = ± 2        [x = -2 is rejected as log is not defined for negative numbers]
⇒ x = 2

Also, 3log_y x = 1
⇒ 3log_y 2 = 1
⇒ log_y 2³ = 1
⇒ 2³ = y¹ 
⇒ y = 8

∴ x + y = 2 + 8 = 10

Hence, option (d).

Given, 
⇒ 1/2, log₄ (2^x - 9) and log₄ (2^x + 17/2) are in AP

⇒ 2 × log₄ (2^x - 9) = 1/2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 + log₄ (2^x + 17/2)

⇒ log₄ (2^x - 9)² = log₄ 2 × (2^x + 17/2)

⇒ (2^x - 9)² = 2 × (2^x + 17/2)

⇒ (a - 9)² = 2a + 17    [Assuming 2^x = a]

⇒ a² - 18a + 81 = 2a + 17

⇒ a² - 20a + 64 = 0

⇒ (a - 16)(a - 4) = 0

⇒ a = 2^x = 16   [4 is rejected as (a - 9) cannot be negative]

Now the first term of the AP = 1/2 and

second term of the AP = log₄ (2^x - 9) = log₄ (16 - 9) = log₄ 7

Common difference = log₄ 7 - 1/2 = log₄ 7 - log₄ 2 = log₄ 7/2

Hence, option (a).