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

Solution:

Since, 1og2 , log(2ⁿ -1) and log (2ⁿ+ 3) are in AP. 

$\begin{array}{ll}\therefore & 2\log \left(2^{\mathrm{n}}-1\right)=\log 2+\log \left(2^{\mathrm{n}}+3\right)\\ \Rightarrow & {\left(2^{\mathrm{n}}-1\right)}^2=2\left(2^{\mathrm{n}}+3\right)\\ \Rightarrow & \left(2^{\mathrm{n}}-5\right)\left(2^{\mathrm{n}}+1\right)=0\end{array}$

As 2ⁿ cannot be negative hence, 

$\begin{array}{l}{2}^{\mathrm{n}}-5=0\\ \Rightarrow 2^{\mathrm{n}}=5\Rightarrow \mathrm{n}={\log }_{2}5\end{array}$