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If $\log_{5} 2, \log_{5} (2^{x} - 5)$ and $\log_{5} (2^{x} - 7 / 2)$ are in A.P., then x is equal to

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Correct option is B

As log5⁡2,log5⁡2x−5, log5⁡2x−7/2 are in A.P., we get 2log5⁡2x−5=log5⁡2+log5⁡2x−7/2⇒ 2x−52=22x−7/2⇒ 2x2−122x+32=0 ⇒ 2x−42x−8=0⇒ 2x=22,23 ⇒ x=2,3.Clearly, x≠2. Therefore x=3

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If log5⁡2,log5⁡2x−5 and log5⁡2x−7/2 are in A.P., then x is equal to

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If $\log_{5} 2, \log_{5} (2^{x} - 5)$ and $\log_{5} (2^{x} - 7 / 2)$ are in A.P., then x is equal to