\(3^x+3^{x+1}+3^{x+2}=117\)
\(\Rightarrow3^x+3^x.3+3^x.9=117\)
\(\Rightarrow3^x.\left(1+3+9\right)=117\)
\(\Rightarrow3^x.13=117\)
\(\Rightarrow3^x=117:13\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(3^x+3^{x+1}+3^{x+2}=117\)
\(\Rightarrow3^x+3^x.3+3^x.9=117\)
\(\Rightarrow3^x.\left(1+3+9\right)=117\)
\(\Rightarrow3^x=117:13\)
\(\Rightarrow3^x=9\)
\(\Leftrightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(3^x+3^{x+1}+3^{x+2}=117\)
=> \(3^x\left(1+3+3^2\right)=117\)
=> \(3^x.13=117\)
=> \(3^x=9\)
mà \(9=3^2\)
=> x = 2
3 x + 3 x+1 + 3 x+2 = 117
3 x + 3 x .3 + 3 x .9 = 117
3 x .(1 + 3 + 9) = 117
3 x .13 = 117
3 x = 117 : 13
3 x = 9 ⇒ 3 x = 3 2
x = 2 x = 2
