If 3 ( 3^x - 2 )\, and ( 3^x + 4 ) are in arithmetic progression, then x is equal to

Answer: Option C

In arithmetic progression common ratio are equal to
$$\log \left(3^x-2\right)-\log 3=$$     $$\log \left(3^x+4\right)-$$   $$\log \left(3^x-2\right)$$
$$\frac{\log \left(3^x-2\right)}{\log 3}=\frac{\log \left(3^x+4\right)}{\log \left(3^x-2\right)}$$     $$\left(\therefore \log a-\log b=\log \frac{a}{b}\right)$$
$$\begin{array}{rl}& \frac{\log 3^x}{\log 2\log 3}=\frac{x\log 3\log 4\log 2}{x\log 3}\\ & \frac{x\log 3}{\log 2\log 3}=\frac{x\log 3\log 4\log 2}{x\log 3}\\ & \frac{x}{\log 2}=\log 4\log 2\\ & x=\log 4\log 2\log 2\\ & x=\log 8\\ & x=\log 2^3\end{array}$$