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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

1 Answer

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(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 logax = \(\frac{1}{n}\) 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 = − \(\frac{1}{3}\).

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