Question

\(3^{x+3}\). \(3^{2x-1}\)\(+\) \(3^{2x}\).\(3^{x+1}\)\(=\)\(324\)

Answer 1

Ta có: \(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)

\(\Leftrightarrow3^{3x+2}+3^{3x+1}=324\)

\(\Leftrightarrow3^{3x+1}\cdot\left(3+1\right)=324\)

\(\Leftrightarrow3^{3x+1}\cdot4=324\)

\(\Leftrightarrow3^{3x+1}=81=3^4\)

\(\Rightarrow3x+1=4\)

\(\Leftrightarrow x=1\)

Answer 2

\(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)   

\(3^{x+3+2x-1}+3^{2x+x+1}=324\)   

\(3^{3x+2}+3^{3x+1}=324\)   

\(3^{3x+1}\cdot\left(3+1\right)=324\)   

\(3^{3x+1}\cdot4=324\)   

\(3^{3x+1}=324:4\)   

\(3^{3x+1}=81\)   

\(3^{3x+1}=3^4\)   

\(\Rightarrow3x+1=4\)   

\(3x=4-1\)   

\(3x=3\)   

\(x=3:3\)   

\(x=1\)