Solve the following equation for x:

log₆ (x - 2) + log₆ (x + 3) = 1

Given,

⇒ log₆ (x - 2) + log₆ (x + 3) = 1

⇒ log₆ (x - 2)(x + 3) = 1

⇒ (x - 2)(x + 3) = 6¹

⇒ (x² + 3x - 2x - 6) = 6

⇒ x² + x - 6 = 6

⇒ x² + x - 12 = 0

⇒ x² + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x - 3)(x + 4) = 0

⇒ x - 3 = 0 or x + 4 = 0

⇒ x = 3 or x = -4.

In this case x ≠ -4 as (x + 3) and (x - 2) will be negative and log of only positive numbers are defined.

Hence, x = 3.