If log 2,;log ({2^n} - 1) and log ({2^n} + 3) | vedclass

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Question

(b) As, $\log 2,\;\log ({2^n} - 1)$ and $\log ({2^n} + 3)$ are in $A.P.$

Therefore, $2\log ({2^n} - 1) = \log 2 + \log ({2^n} + 3)$

$ \Rightarro

Vedclass · Accepted Answer

${\log _2}5$

Vedclass · Answer

(b) As, $\log 2,\;\log ({2^n} - 1)$ and $\log ({2^n} + 3)$ are in $A.P.$

Therefore, $2\log ({2^n} - 1) = \log 2 + \log ({2^n} + 3)$

$ \Rightarrow ({2^n} - 5)({2^n} + 1) = 0$

As ${2^n}$ cannot be negative, hence ${2^n} - 5 = 0$

$ \Rightarrow {2^n} = 5$ or $n = {\log _2}5$.

If $\log 2,\;\log ({2^n} - 1)$ and $\log ({2^n} + 3)$ are in $A.P.$, then $n =$

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