The number of solution of log4 (x – 1) = log2 (x – 3) is :(a) 0 (

1.	The number of solution of log4 (x – 1) = log2 (x – 3) is :(a) 0 (b) 5 (c) 2 (d) 3
Answer» (b) 5 log₄ (x – 1) = log₂ (x – 3) ⇒ log_2² (x – 1) log₂ (x – 3) ⇒ \(\frac{1}{2}\) log₂(x – 1) log₂ (x – 3) \(\bigg[\)Using log_mⁿ (x) = \(\frac{1}{n}\) log_mx\(\bigg]\) ⇒ log₂ (x – 1) = 2 log₂ (x – 3) ⇒ log₂ (x – 1) = log₂ (x – 3)² ⇒ (x – 1) = (x – 3)² ⇒ (x – 1) = x² – 6x + 9 ⇒ x² – 7x + 10 = 0 ⇒ (x – 2) (x – 5) = 0 ⇒ x = 2 or 5. x = 2 is inadmissible as log₂ (x – 3) is not defined when x = 2. ∴ x = 5.

The number of solution of log4 (x – 1) = log2 (x – 3) is :(a) 0 (b) 5 (c) 2 (d) 3

Answer»

(b) 5

log₄ (x – 1) = log₂ (x – 3)

⇒ log_2² (x – 1) log₂ (x – 3)

⇒ \(\frac{1}{2}\) log₂(x – 1) log₂ (x – 3) \(\bigg[\)Using log_mⁿ (x) = \(\frac{1}{n}\) log_mx\(\bigg]\)

⇒ log₂ (x – 1) = 2 log₂ (x – 3)

⇒ log₂ (x – 1) = log₂ (x – 3)²

⇒ (x – 1) = (x – 3)² ⇒ (x – 1) = x² – 6x + 9

⇒ x² – 7x + 10 = 0 ⇒ (x – 2) (x – 5) = 0 ⇒ x = 2 or 5.

x = 2 is inadmissible as log₂ (x – 3) is not defined when x = 2.

∴ x = 5.