Question

If a is a perfect square, what is the value of a, given that a^x = 1/8 and x = 3/2[(log₂ a) - 3] ? 
1. 2
2. 4
3. 16
4. 64

Answer 1

Correct Answer - Option 2 : 4

Calculation :

Given, a^x = 1/8, x = 3/2[(log2 a) - 3]

⇒ a^{3/2[(log₂ a) - 3]} = 1/8

Applying log to base 2 in both sides, we get,

3/2[(log₂ a) - 3] log₂ a = log₂ (1/8)

⇒ 3/2[(log2 a) - 3] log2 a = -3

Let log₂ a = x

⇒ (x - 3) x = -2

⇒ x² -3x + 2 = 0

⇒ (x - 1)(x - 2) = 0

⇒ x = 1 or 2

⇒ log₂ a = 1 or log₂ a = 2

⇒ a = 2 or a = 4

As given 'a' is a perfect square ⇒ a = 4

Whenever there is a quadratic equation in logarithm, always check whether the final answer satisfies the given condition in the question and logarithmic properties.