2.16 ≥ \(2^n\) ≥ 4

⇒ 32 ≥ \(2^n\) ≥4

⇒ \(2^5\) ≥ \(2^n\) ≥ \(2^2\)

⇒ 5 ≥ n ≥ 2

⇒ n ∈ {5;4;3;2}

a).

\(2.16=2.2^4=2^5\\ 4=2^2\)

theo đề bài, ta có: \(2^5\ge2^n>2^2\Rightarrow5\ge n>2\)

vì n là số tự nhiên nên : \(n=5;4;3\)

b).

\(9.27=3^2.3^3=3^5\\ 243=3^5\)

theo đề bài, ta có: \(3^5\le3^n\le3^5\Rightarrow5\le n\le5\)

=> n=5

Giải:

a)2.16\(\ge\)2ⁿ>4

2.2⁴\(\ge\)2ⁿ>2²

2⁵\(\ge\)2ⁿ>2²

\(\Rightarrow\)5\(\ge\)n>2

\(\Rightarrow\)n\(\in\){3;4;5}

b)9.27\(\le\)3ⁿ\(\le\)243

3².3³\(\le\)3ⁿ\(\le\)3⁵

3⁵\(\le\)3ⁿ\(\le\)3⁵

5\(\le\)n\(\le\)5

\(\Rightarrow\)n=5

a) 2.16\(\ge\)2ⁿ>4

=>2.2⁴\(\ge\)2ⁿ>4

=>2^5\(\ge\)2ⁿ>2²

^{=>5\(\ge\)n>2}

^=>n\(\in\){3,4,5}

b)9.27\(\le\)3ⁿ\(\le\)243

=>3².3³\(\le\)3ⁿ\(\le\)3⁵

=>3⁵\(\le\)3ⁿ\(\le\)3⁵

=>5^\(\le\)n\(\le\)5

=>n=5

2.16 ≥ 2n > 4 ⇒ 2. 24 ≥ 2n > 22

⇒ 25 ≥ 2n > 22

⇒ 5 ≥ n > 2

⇒ n ∈ {3; 4; 5}

Đáp án B

a, \(2.16\ge2^n>4\Rightarrow2^5\ge2^n>2^2\Rightarrow5\ge n>2\Rightarrow n\in\left\{3;4;5\right\}\)

b,\(9.27\le3^n\le243\Rightarrow3^5\le3^n\le3^5\Rightarrow n=5\)

\(2.32\ge2^n>8\\ \Rightarrow2^6\ge2^n>2^3\\ \Rightarrow n\in\left\{4;5;6\right\}\)

\(2.32=2.2^5=2^6\ge2^n>8=2^3\)

Do \(n\in N\)

\(\Rightarrow n\in\left\{6;5;4\right\}\)

a) ta có 2.16\(\ge\)2ⁿ > 4

    \(\rightarrow\)2.2⁴\(\ge\)2ⁿ>2²

    ^{\(\rightarrow\)} 2⁵\(\ge\)2ⁿ>2²

^{\(\Rightarrow\)} n\(\in\){ 3;4;5}

b) làm tương tự

a. 2.16 \(\ge\)2ⁿ>4

=> 32 \(\ge\)2ⁿ>4

=> 2⁵ \(\ge\)2ⁿ>2²

=> n \(\in\left\{3;4;5\right\}\)

b. \(9.27\le3^n\le243\)

=> \(243\le3^n\le243\)

=> \(3^5\le3^n\le3^5\)

=> n=5