If log 2, log

The numbers $\log 2, \log \left(2^{x}-1\right)$ and $\log \left(2^{x}+3\right)$ are in A.P.

$\therefore \log \left(2^{x}-1\right)-\log 2=\log \left(2^{x}+3\right)-\log \left(2^{x}-1\right)$

$\Rightarrow \log \left(\frac{2^{x}-1}{2}\right)=\log \left(\frac{2^{x}+3}{2^{x}-1}\right)$

$\Rightarrow \frac{2^{x}-1}{2}=\frac{2^{x}+3}{2^{x}-1}$

$\Rightarrow\left(2^{x}-1\right)^{2}=2\left(2^{x}+3\right)$

$\Rightarrow 2^{2 x}+1-2.2^{x}=2.2^{x}+6$

$\Rightarrow 2^{2 x}-4.2^{x}-5=0$

Let $2^{x}=y$.

$\Rightarrow y^{2}-4 y-5=0$

$\Rightarrow(y-5)(y+1)=0$

$\Rightarrow y=5$ or $y=-1$

$\therefore 2^{x}=5$ or $2^{x}=-1$ (not possible)

Taking log on both the sides, we get:

$\log 2^{x}=\log 5$

$\Rightarrow x \log 2=\log 5$

$\Rightarrow x=\frac{\log 5}{\log 2}=\log _{2} 5$

$\Rightarrow x=\log _{2} 5$

Disclaimer: The question in the book has some error, so, this solution is not matching with the solution given in the book. This solution here is created according to the question given in the book.