If ${\log }_{10}2,{\log }_{10}\left(2^x-1\right)$ and ${\log }_{10}\left(2^x+3\right)$ are three consecutive terms of an A.P. for

 

$2,2^x-1,2^x+3$ are in G.P.

$\begin{array}{l}\Rightarrow {\left(2^x-1\right)}^2=2\left(2^x+3\right)\\ \Rightarrow 2^{2x}-2\left(2^x\right)+1=2\left(2^x\right)+6\\ \Rightarrow 2^{2x}-4\left(2^x\right)-5=0\\ \Rightarrow \left(2^x-5\right)(2x+1)=0\\ \Rightarrow 2^x=5\left[\because 2^x+1>0\right]\end{array}$

Thus, there is exactly one value of $x.$