\(A=3+3^2+3^3+......+3^{2017}\)
\(\Leftrightarrow3A=3^2+3^3+3^4......+3^{2018}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+......+3^{2018}\right)-\left(3+3^2+3^3+......+3^{2017}\right)\)
\(\Leftrightarrow2A=3^{2018}-3\)
\(\Leftrightarrow2A+3=3^{2018}=3^x\)
\(\Rightarrow x=2018\)
Vậy \(x=2018\)