Test: Logarithm- 2 - CAT MCQ

Converting all logs to base 10.

⇒ log_a32 + log_a15 = 4

⇒ log_a(32 × 15) = 4

⇒ a⁴ = 480

This is possible when 4 < a < 5.

Hence, option (a).

Taking log on both sides (Choosing the base to be 3)

⇒ y² × log₃5 × log₃2 = log₂3 × log₃5

⇒ y² × log₃2 = log₂3

⇒ - y² = (log₂3)²

⇒ y = - log₂3 (∵ y is a negative number)

⇒ y = log₂(1/3)

Hence, option (d).

We know logab is positive and sum of a positive number and its reciprocal is always ≥ 2, i.e.,

∴ log_ab + log_ba ≥ 2

⇒ 2 – (log_ab + log_ba) ≤ 2 – 2

⇒ 2 – (log_ab + log_ba) ≤ 0

∴ 2 – (log_ab + log_ba) cannot take positive value i.e., it cannot be equal to 1 (option (b))

Hence, option (c).

Since all the 4 options have A + B, lets add A and B.

∴ A + B = log_a30 + (-log_a(5/3)) 

= log_a30 – log_a(5/3) = log_a(30 × 3/5) 

= log_a(18) = log_a(2 × 3²)

⇒ A + B = log_a(2) + 2log_a(3)

⇒ A + B = 3 + 2log_a(3) [∵ log₂a = 1/3 ∴ log_a2 = 3]

⇒ log_a(3) = (A + B - 3)/2

⇒ log₃(a) = 2/(A + B - 3)

Hence, option (c).

Given: 2^6x + 2^3x+2 - 21 = 0

Replace 2^3x with y

So, 2^6x = y²

Now, 2^6x + 2^3x+2 - 21 = 0 can be rewritten as y² + 2² × 2^3x - 21 = 0

y² + 4y - 21 = 0

Solving the above quadratic equation,

(y + 7) (y - 3) = 0

So, y = -7 or +3

y = - 7 is rejected (since, y = 2^3x which should always be positive)

⇒ 2^3x = 3

Taking log on both sides,

log₂⁡3 = 3x 

Hence, option (a).