\(A=3+3^2+3^3+.....+3^{2006}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+.......+3^{2007}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+.......+3^{2007}\right)-\left(3+3^2+......+3^{2006}\right)\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow2A+3=3^{2007}\)
Mà \(2A+3=3^x\)
\(\Leftrightarrow3^{2007}=3^x\)
\(\Leftrightarrow x=2007\)
Vậy....