If ${\log }_{3}2,{\log }_3(2^x-5)$  and  ${\log }_3(2^x-\frac{7}{2})$ are in  $\text{AP},x$ is equal to

$\because {\log }_{3}2,{\log }_3(2^x-5)$  and ${\log }_3(2^x-\frac{7}{2})$  are in AP  ....(i)
For defined, $2^x-5>0$  and $2^x-\frac{7}{2}>0$ 
  $\therefore 2^x>5$      ....(ii)
From Eq. (i), $2,2^x-5,2^x-\frac{7}{2}$  are in GP
 $\begin{array}{ll}\therefore & {(2^x-5)}^2=2·(2^x-\frac{7}{2})\\ \Rightarrow & 2^{2x}-12·2^x+32=0\\ \Rightarrow & (2^x-8)(2^x-4)=0\\ \therefore & 2^x=8,4\\ \Rightarrow & 2^x=8=2^3,2^x\ne 4[\text{fromEq}.(\text{ii}\left)\right]\\ \therefore & x=3\end{array}$