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