\(9< 3^n< 27\)

\(3^2< 3^n< 3^3\)

ko có số nguyên dương nào thỏa mãn điều kiện

ta có: 9<3^n<27 nên 3^2<3^n<3^3

Suy ra n không có giá trị thỏa mãn 9<3^n<27

Ta có :

\(10\le n\le99\)

\(\Rightarrow21\le2n+1\le201\)

\(\Rightarrow2n+1\) là số chính phương lẻ (1)

\(\Rightarrow2n+1\in\left\{25;49;81;121;169\right\}\)

\(\Rightarrow n\in\left\{12;24;40;60;84\right\}\)

\(\Rightarrow3n+1\in\left\{37;73;121;181;253\right\}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{2n+1}{3n+1}=\dfrac{2.40+1}{3.40+1}=\dfrac{81}{121}=\left(\dfrac{9}{11}\right)^2\left(n=40\right)\)

\(\Rightarrow dpcm\)

\(\Rightarrow n=40⋮40\Rightarrow dpcm\)

\(3^{3n+1}=9^{n+2}\)

\(3^{3n+1}=3^{2.\left(n+2\right)}\)

\(3^{3n+1}=3^{2n+4}\)

=> 3n + 1 = 2n + 4

=> 3n - 2n = 4 - 1

=> n = 3 

3³ⁿ⁺¹ = 9ⁿ⁺²

3³ⁿ⁺¹ = 3²⁽ⁿ⁺²⁾

3³ⁿ⁺¹ = 3²ⁿ⁺⁴

3n + 1 = 2n + 4

2n - 3n = 1 - 4

-n = -3

n = 3 

\(3^{3n+1}=9^{n+2}=\left(3^2\right)^{2n+2}=2^{4n+4}=>3n+1=4n+4=>n=-3\)

 \(3^{3n+1}=9^{n+2}\Rightarrow3^{3n+1}=\left(3^2\right)^{n+2}\)

\(\Rightarrow3^{3n+1}=3^{2\left(n+2\right)}\Rightarrow3n+1=2\left(n+2\right)\)

\(\Rightarrow3n+1=2n+4\Rightarrow3n-2n=4-1\)

\(\Rightarrow n=3\)