Ta có: \(A=3+3^2+3^3+...+3^{2008}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2009}\)
Trừ \(3A-A=3^2+3^3+3^4+...+3^{2009}-3-3^2-3^3-...-3^{2008}\)
\(\Rightarrow2A=3^{2009}-3\)
Mà \(2A=3^x-3\)
\(\Rightarrow3^x=3^{2009}\)
\(\Rightarrow x=2009.\)
Vậy x = 2009.
\(a=3+3^2+3^3+...+3^{2008}\)
\(3a=3^2+3^3+3^4+...+3^{2009}\)
\(3a-a=\left(3^2+3^3+3^4+...+3^{2009}\right)-\left(3+3^2+3^3+...+3^{2008}\right)\)
\(2a=3^{2009}-3\)
\(2a+3=3^{2009}=3^x\)
\(x=2009\)
\(A=3+3^2+3^3+......+3^{2008}\)
\(\Leftrightarrow3A=3^2+3^3+......+3^{2009}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4......+3^{2009}\right)-\left(3+3^2+3^3+......+3^{2008}\right)\)
\(\Leftrightarrow2A=3^{2009}-3\)
\(\Leftrightarrow2A+3=3^{2009}=3^x\)
Vậy \(x=2009\) để \(2A+3=3^x\)
\(a=3+3^2+3^3+...+3^{2008}\)
\(3a=3^2+3^3+3^4+...+3^{2009}\)
\(3a-a=\left(3^2+3^3+3^4+...+3^{2009}\right)-\left(3+3^2+3^3+...+3^{2008}\right)\)
\(2a=3^{2009}-3\)
\(2a+3=3^x=3^{2009}-3+3=3^{2009}\)
\(x=2009\)