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If $\log 2, \log (2^{n} - 1)$ and $(2^{n} + 3)$ are in AP, then n is equal to

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Correct option is B

Since, 1og2 , log(2n -1) and log (2n+ 3) are in AP. ∴2log⁡2n−1=log⁡2+log⁡2n+3⇒2n−12=22n+3⇒2n−52n+1=0As 2n cannot be negative hence, 2n−5=0⇒ 2n=5⇒n=log2⁡5

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If log ⁡2,log ⁡2n−1 and 2n+3 are in AP, then n is equal to

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If $\log 2, \log (2^{n} - 1)$ and $(2^{n} + 3)$ are in AP, then n is equal to