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Em kiểm tra lại đề bài, \(3^x+3^{x+2}+3^{x+2}\) hay \(3^x+3^{x+1}+3^{x+2}\)
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$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
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\(=3^{x+1}\left(1+3+3^2\right)+...+3^{x+10}\left(1+3+3^2\right)=\)
\(=3^x.3.13+...+3^{x+9}.3.13=\)
\(39\left(3^x+...+3^{x+9}\right)⋮39\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(3^x+3^{x+2}+3^{x+3}=3^5\cdot37\)
\(\Leftrightarrow3^x\cdot\left(1+9+27\right)=3^5\cdot37\)
hay x=5
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\(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x+3^x.3+3^x.3^2=117\)
\(3^x\left(1+3+3^2\right)=117\)
\(3^x.13=117\)
\(3^x=9\)
\(\Rightarrow x=2\)
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a) Ta có 2x + 2x + 1 + 2x + 2 = 56
⇒ 2x ( 1 + 21 + 22 ) = 56
⇒ 2x . 7 = 56
⇒ 2x = 56 : 7 = 8 = 23
Vậy x = 3
b) Ta có 3x + 3x + 2 + 3x + 3 = 111
⇒ 3x ( 1 + 32 + 33 ) = 111
⇒ 3x . 37 = 111
⇒ 3x = 111 : 37 = 3 = 31
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(3x+3x+1+3x+2+3x+3=29160\)
\(\Leftrightarrow12x=29154\)
hay x=2429,5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x+3x+1+3x+2+3x+3=29160\Rightarrow3x+3x+3x+3x=29160-1-2-3\)
\(\Rightarrow12x=29154\Rightarrow x=\dfrac{4859}{2}=2429,5\)
\(3^x.3^{x+1}.3^{x+2}=81.3^5\)
=> \(\left(3^x\right)^3.\left(3^1.3^2\right)=3^9\)
=> \(\left(3^x\right)^3.3^3=3^9\)
=> \(3^x.3^3.3^3=3^9\)
=> \(3^x=3^9:\left(3^{3+3}\right)\)
=> \(3^x=3^3\)
=> \(x=3\)