The Solution set of the given equation -

Asked by thebluebloo | 7th Apr, 2009, 06:59: PM

Expert Answer:

logx2log2x2=log4x2

logx2log2x2=log4x2

⇒ (log2/ logx) Χ (log2/log2x) = (log2/log4x)

⇒ log 2. log4x = logx. log2x

⇒ log2 (log4 + logx) = logx (log2 + logx)

⇒ log2. log4 = (log x)2

⇒ 2 (log2)2 = (log x)2

log x = ± 2 . log2

⇒ x = e±2log2

⇒ x  = 2and 2-2

Hence the correct option is (1)

Answered by  | 20th May, 2009, 03:44: PM

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