If 2, ; ( 2^n - 1) and ( 2^n + 3) are in A.P., then n = - Vidyadip

MCQ

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

A
$5/2$
✓
${\log _2}5$
C
${\log _3}5$
D
$3/2$

✓

Answer

Correct option: B.

${\log _2}5$

b
(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 $n = {\log _2}5$.

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