\(3^{x+3}.3^{2x-1}.3^x=729\)
\(3^{x+3+2x-1+x}=729\)
\(3^{4x+2}=3^6\)
\(\Rightarrow4x+2=6\)
\(\Rightarrow4x=4\)
\(\Rightarrow x=1\)
\(3^{x+3}.3^{2x-1}.3^x=729\)
\(\Leftrightarrow3^{x+3+2x-1+x}=3^6\)
\(\Leftrightarrow3^{4x+2}=6\)
\(\Leftrightarrow4x+2=6\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
\(3^{x+3}.3^{2x-1}.3x=729\)
\(3^{x+3+2x-1+x}=3^6\)
\(3^{4x+2}=6\)
\(4x+2=6\)
\(4x=4\)
\(\Rightarrow x=1\)
\(3^{x+3}\cdot3^{2x-1}\cdot3^x=729→3^{4x+2}=3^6→4x+2=6→x=1\)
\(3^{x+3}\cdot3^{2x-1}\cdot3^x=729\)
\(3^{x+3+2x-1+x}=3^6\)
\(\Rightarrow4x+2=6\) ( cộng mũ vô nha mình làm gọn 1 bước )
\(4x=6-2\)
\(4x=4\)
\(x=4:4\)
\(x=1\)