(d) 1
Given, log2(3x-2) = \(\text{log}_{\frac{1}{2}}\)x ⇒ log2(3x-2) = log2-1 x
⇒ log2(3x-2) = \(\frac{1}{-1}\) log2 x \(\bigg[\)Using logan x = \(\frac{1}{n}\) loga x\(\bigg]\)
⇒ log2 (3x – 2) = (– 1) log2 x = log2 x– 1 [Using n loga x = loga 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 = − \(\frac{1}{3}\).