\(3^{2x+4}:3^{x+1}=81\)
\(3^{2x+4-x-1}=3^4\)
\(3^{x+3}=3^4\)
\(\Rightarrow x+3=4\)
\(\Rightarrow x=1\)
\(3^{2x+4}:3^{x+2}=81\)
\(\Rightarrow3^{2x+4-x-2}=3^4\)
\(\Rightarrow3^{x+2}=3^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=4-2=2\)
Nhầm đề :v
\(3^{2x+4}:3^{x+2}=81\)
\(3^{2x+4-x-2}=3^4\)
\(3^{x+2}=3^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=2\)
32n+4:3x+1 =81
-> 32n+4:3x+1=34
-> 2n+4-x-1=4
-> x+3=4
-> x=1
Vậy x=1
\(3^{2x+4}\):\(3^{x+2}\)=81
\(3^{\left(2x+4\right)-\left(x+2\right)}\)=81
\(3^{2x+4-x-2}\)=81
\(3^{\left(2x-x\right)+\left(4-2\right)}\)=81
\(3^{x+2}\)=81
\(3^{x+2}\)=\(3^4\)
\(x+2=4\)
\(x=4-2\)
\(x=2\)