If $$\log 3\log \left( {{3^x} - 2} \right)\,$$   and $$\log \left( {{3^x} + 4} \right)$$   are in arithmetic progression, then x is equal to

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