<=> 3².3³ ≤ 3ⁿ ≤ 3⁵

<=> 3⁵ ≤ 3ⁿ ≤ 3⁵

=> n = 5

a) \(2\cdot16\ge2^n>4\\ 2\cdot2^4\ge2^n>2^2\\ 2^5\ge2^n>2^2\\ \Rightarrow2^n\in\left\{2^3;2^4;2^5\right\}\\ \Rightarrow n\in\left\{3;4;5\right\}\)Vậy \(n\in\left\{3;4;5\right\}\)

b) \(9\cdot27\le3^n\le243\\ 3^2\cdot3^3\le3^n\le3^5\\ 3^5\le3^n\le3^5\\ \Rightarrow3^n=3^5\\ \Rightarrow n=5\)Vậy n = 5

a) \(2.16\ge2^n>4\)

\(\Rightarrow32\ge2^n>4\)

\(\Rightarrow2^5\ge2^n>2^2\)

\(\Rightarrow5\ge n>2\)

\(\Rightarrow\left\{{}\begin{matrix}n=3\\n=4\\n=5\end{matrix}\right.\)

Vậy \(n\in\left\{3;4;5\right\}.\)

b) \(9.27\le3^n\le243\)

\(\Rightarrow243\le3^n\le243\)

\(\Rightarrow3^5\le3^n\le3^5\)

\(\Rightarrow5\le n\le5\)

\(\Rightarrow n=5\)

Vậy \(n=5.\)

Chúc bạn học tốt!

Thắng Nguyễn alibaba nguyễn Hoàng Lê Bảo Ngọc  help me

a) \(\frac{1}{9}\cdot27^n=3^n\)

\(\frac{1}{9}=\frac{3^n}{27^n}\)

\(\frac{1}{9}=\frac{3^n}{3^{3n}}\)

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

=> 3²ⁿ = 9 = 3²

=> 2n = 2

=> n = 1

9.27 ≤ 3n ≤ 243 ⇒ 32.33 ≤ 3n ≤ 35

⇒ 35 ≤ 3n ≤ 35 ⇒ n = 5

1, 32 < 2^n < 128

    2^5 < 2^n < 2^7

=> 5 < n < 7 

Vì n là nguyên dương => n = 6 

2,  2.16 > (=) 2^n > 4 

    2.2^4 > (=) 2^n > 2^2 

  2^5 > (=) 2^n  > 2^2

 5 >(=) n > 2 => n = 5 ; 4 ; 3 

3, 9.27 < 3^n <= 243

  3^2 . 3^3 < 3^n <= 3^5

     3^5       < 3^n  <=5

   5 < n <= 5 ( không có n)      

\(9.27\le\left(\frac{1}{3}\right)^x\le27.243\)

\(\Rightarrow243\le\left(\frac{1}{3}\right)^x\le6561\)

\(\Rightarrow\left(\frac{1}{3}\right)^{-5}\le\left(\frac{1}{3}\right)^x\le\left(\frac{1}{3}\right)^{-8}\)

\(\Rightarrow-5\le x\le-8\)

\(\Rightarrow\left\{{}\begin{matrix}x=-5\\x=-6\\x=-7\\x=-8\end{matrix}\right.\)

Vậy \(x\in\left\{-5;-6;-7;-8\right\}.\)

Chúc bạn học tốt!

\(9.27\le\left(\frac{1}{3}\right)^x\le27.243\)

\(243\le\left(\frac{1}{3}\right)^x\le6561\)

đến đây toi tịt == 

à để xem !!! như thế này này :

\(3^5\le\left(\frac{1}{3}\right)^x\le3^8\)

Ta có \(\frac{1}{3}=0,\left(3\right)=\sqrt{3}=3\)( bt là số vô tỉ nx xem đã )

\(\Rightarrow3^5\le3^x\le3^8\)

\(\Rightarrow x=6;7\)

\(9.27\le\left(\frac{1}{3}\right)^x\le27.243\)

\(\Rightarrow3^2.3^3\le\left(\frac{1}{3}\right)^x\le3^3.3^5\)

\(\Rightarrow3^5\le\left(\frac{1}{3}\right)^x\le3^8\)

\(\Rightarrow x\in\left\{-5;-6;-7;-8\right\}\)