If log 2, (\log(2^{n} - 1)) and (\log(2^{n} + 3)) are in A.P... | MATH - ARITHMETIC PROGRESSION | SolveeIt

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

5/2

$\log_{2}5$

$\log_{3}5$

$\frac{3}{2}$

Answer

$\log_{2}5$

Explanation

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

$2\log(2^{n} - 1) = \log 2 + \log(2^{n} + 3)$ ⇒ $(2^{n} - 5)(2^{n} + 1) = 0$

As $2^{n}$ cannot be negative, hence $2^{n} - 5 = 0$ ⇒ $2^{n} = 5$ or

$n = \log_{2}5$

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