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

If $$log{3}, log(3^{x} - 2)$$ and $$log (3^{x}+ 4)$$ are in arithmetic progression
Then, $$2*log(3^{x} - 2) = log{3}+log (3^{x}+ 4)$$
Thus, $$log{(3^{x} - 2)^2} = log{3(3^x+4)}$$
Thus, $$(3^{x} - 2)^2 = 3(3^x+4)$$
=> $$3^{2x} - 4*3^x +4 = 3*3^x + 12$$
=> $$3^{2x} - 7*3^x - 8 = 0$$
=> $$(3^x+1)*(3^x-8) = 0$$
But $$3^x+1 \neq 0$$
Thus, $$3^x = 8$$
Hence, $$x = log_{3}{8}$$
Hence, option B is the correct answer.