The value of x satisfying log2(3x – 2) = \(\text{log}_{\frac{1}{2}}\

1.	The value of x satisfying log2(3x – 2) = \(\text{log}_{\frac{1}{2}}\) x is(a) – 1 (b) − \(\frac{1}{3}\)(c) \(\frac{1}{3}\)(d) 1
Answer» (d) 1 Given, log₂(3x-2) = \(\text{log}_{\frac{1}{2}}\)x ⇒ log₂(3x-2) = log_2^-1 x ⇒ log₂(3x-2) = \(\frac{1}{-1}\) log₂ x \(\bigg[\)Using log_aⁿx = \(\frac{1}{n}\) log_a x\(\bigg]\) ⇒ log₂ (3x – 2) = (– 1) log₂ x = log₂ x^{– 1}[Using n log_a x = log_a xⁿ] ⇒ (3x – 2) = x^{– 1} = \(\frac{1}{x}\) ⇒ 3x² – 2x – 1 = 0 ⇒ (3x + 1) (x – 1) = 0 ⇒ \(-\frac{1}{3}\) x = or 1 ⇒ x = 1, since log₂ (3x – 2) is not defined when x = − \(\frac{1}{3}\).

The value of x satisfying log2(3x – 2) = \(\text{log}_{\frac{1}{2}}\) x is(a) – 1 (b) − \(\frac{1}{3}\)(c) \(\frac{1}{3}\)(d) 1

Answer»

(d) 1

Given, log₂(3x-2) = \(\text{log}_{\frac{1}{2}}\)x ⇒ log₂(3x-2) = log_2^-1 x

⇒ log₂(3x-2) = \(\frac{1}{-1}\) log₂ x \(\bigg[\)Using log_aⁿx = \(\frac{1}{n}\) log_a x\(\bigg]\)

⇒ log₂ (3x – 2) = (– 1) log₂ x = log₂ x^{– 1}[Using n log_a x = log_a xⁿ]

⇒ (3x – 2) = x^{– 1} = \(\frac{1}{x}\) ⇒ 3x² – 2x – 1 = 0

⇒ (3x + 1) (x – 1) = 0 ⇒ \(-\frac{1}{3}\) x = or 1

⇒ x = 1, since log₂ (3x – 2) is not defined when x = − \(\frac{1}{3}\).