1. Let $a=\log_3 2,\quad b=\log_3(2^x-5),\quad c=\log_3\left(2^x-\frac72\right).$

   Since these are in arithmetic progression, the middle term is the average of the other two: $2b=a+c.$

2. Substitute the logarithms: $2\log_3(2^x-5)=\log_3 2+\log_3\left(2^x-\frac72\right).$

3. Use log properties: $\log_3(2^x-5)^2=\log_3\left[2\left(2^x-\frac72\right)\right].$

   Hence, $(2^x-5)^2=2\left(2^x-\frac72\right)=2^{x+1}-7.$

4. Let $y=2^x.$ Then the equation becomes $(y-5)^2=2y-7.$

   Expanding, $y^2-10y+25=2y-7.$ $y^2-12y+32=0.$

5. Factorize: $(y-4)(y-8)=0.$ So, $y=4\quad \text{or} \quad y=8.$

6. Now check domain restrictions of the logarithms:

   • $2^x-5>0 \Rightarrow y>5$
   • $2^x-\frac72>0 \Rightarrow y>\frac72$

   So we must have $y>5$. Thus $y=4$ is invalid, and only $y=8$ is valid.

7. Therefore, $2^x=8=2^3 \Rightarrow x=3.$

So the required integer value is $3$.