\(3^n.3^1=3^3\)
\(3^n=3^3:3^1\)
\(3^n=3^{3-1}=3^2\Rightarrow n=2\)
tích nha
\(3^{n+1}=27\)
\(\Rightarrow3^{n+1}=3^3\)
\(\Rightarrow n+1=3\)
\(\Rightarrow n=2\)
K NHÉ BẠN
\(3^{n+1}=27\)
\(3^{n+1}=3^3\)
\(\Rightarrow n+1=3\)
\(n=3-1\)
\(n=2\)
Vậy n=2
k mk nha!
\(3^{n+1}=27\)
\(3^{n+1}=3^3\)
\(=>n+1=3\)
\(n=3-1\)
\(n=2\)