3log8x=log4(x+6)

1.	3log8x=log4(x+6)
Answer» 3 log₈x = log₄ (x+6) ⇒ 3 log_2³x = log_2² (x+6) (∵ log_a^mb = 1/m log_ab) ⇒ 3/3 log₂x = 1/2 log₂(x+6) ⇒ 2 log₂x = log₂ (x+6) ⇒ log₂x² = log₂ (x+6) (∵ n log a = log aⁿ) ⇒ x² = x+6 (By taking anti log) ⇒ x² - x - 6 = 0 ⇒ x²- 3x + 2x - 6 = 0 ⇒ x(x-3) + 2(x-3) = 0 ⇒ (x+2) (x-3) = 0 ⇒ x+2 = 0 or x-3 = 0 ⇒ x = -2(Not possible) or x = 3) (∵ Domain of log x is x > 0) Hence, solution is x = 3

Answer»

3 log₈x = log₄ (x+6)

⇒ 3 log_2³x = log_2² (x+6) (∵ log_a^mb = 1/m log_ab)

⇒ 3/3 log₂x = 1/2 log₂(x+6)

⇒ 2 log₂x = log₂ (x+6)

⇒ log₂x² = log₂ (x+6) (∵ n log a = log aⁿ)

⇒ x² = x+6 (By taking anti log)

⇒ x² - x - 6 = 0

⇒ x²- 3x + 2x - 6 = 0

⇒ x(x-3) + 2(x-3) = 0

⇒ (x+2) (x-3) = 0

⇒ x+2 = 0 or x-3 = 0

⇒ x = -2(Not possible) or x = 3)

(∵ Domain of log x is x > 0)

Hence, solution is x = 3

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