First slide

Arithmetic progression

Question

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

Easy

A

$\frac{5}{2}$
B

$\log_{2} 5$
C

$\log_{3} 5$
D

$\frac{3}{2}$

Solution

Since, 1og2 , log(2ⁿ-1) and log (2ⁿ+ 3) are in AP.
$\begin{array}{ll} ∴ & 2 \log (2^{n} - 1) = \log 2 + \log (2^{n} + 3) \\ \Rightarrow & {(2^{n} - 1)}^{2} = 2 (2^{n} + 3) \\ \Rightarrow & (2^{n} - 5) (2^{n} + 1) = 0 \end{array}$
As 2ⁿ cannot be negative hence,
$\begin{array}{l} 2^{n} - 5 = 0 \\ \Rightarrow 2^{n} = 5 \Rightarrow n = \log_{2} 5 \end{array}$

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