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

Question

Dr. Abubacker · Accepted Answer

(b)
Since, $\log{2,\log{(2^{n} - 1)}and\log{(2^{n} + 3)}\ }$are in AP.
$	herefore\ 2\log{\left( 2^{n} - 1 ight) = \log{2 + \log{(2^{n} + 3)}}}$
$\Rightarrow \ \ \left( 2^{n} - 1 ight)^{2} = 2(2^{n} + 3)$
$\Rightarrow \ \ \left( 2^{n} - 5 ight)\left( 2^{n} + 1 ight) = 0$
As $2^{n}$ cannot be negative hence, $2^{n} - 5 = 0$
$\Rightarrow \ \ 2^{n} = 5 \Rightarrow n = {log_{2}}5$

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

Explanation

Reviewed by Dr. Abubacker

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