a) \(\dfrac{1}{8}.16^n=2^n\)

=>\(\dfrac{2^{4n}}{2^3}=2^n\)

=>\(2^{4n-3}=2^n\)

=>4n-3=n

=>3n-3=0

=>n=1.

b) \(27< 3^n< 243\)

=>\(3^3< 3^n< 3^5\). Mà n là số tự nhiên.

- Vậy n=4

a) \(\dfrac{1}{8}.16^n=2^n\)

\(\Rightarrow2^{4n}=2^3.2^n\)

\(\Rightarrow4n=3+n\)

\(\Rightarrow3n=3\)

\(\Rightarrow n=1\)

Vậy: \(n=1\)

b) \(27< 3^n< 243\)

\(\Rightarrow3^3< 3^n< 3^5\)

\(\Rightarrow3< n< 5\)

\(\Rightarrow n=4\)

Vậy: \(n=4\)

mk chắc chắn 100% là n >1

b)\(27< 3^n< 243\)

\(3^3< 3^n< 3^5\)

\(\Rightarrow3< n< 5\)

\(\Rightarrow n\in\left\{4\right\}\)

a)\(\frac{1}{8}.16^n=2^n\)

\(\frac{16^n}{8}=2^n\)

\(\Rightarrow\frac{2^{4n}}{2^3}=2^n\)

\(\Rightarrow2^{4n-3}=2^n\)

\(\Rightarrow4n-3=n\)

\(\Rightarrow n=1\)

ok con de

23

Bạn ghi cách làm luôn hộ mk

a=1

b=4

a. 1/8=2ⁿ:16ⁿ

1/8=1/8ⁿ

=>n=1

b.27<3ⁿ<243

<=>3³<3ⁿ<3⁵

=>n=4

a) 1/8 . 16ⁿ = 2ⁿ

    1/8          = 2ⁿ : 16ⁿ

    1/8          = ( 2/16 )ⁿ

    1/8          = ( 1/8 )ⁿ

=> n = 1

b) 27 < 3ⁿ < 243

    3³ < 3ⁿ < 3⁵

=> n = 4

\(\frac{1}{8}.16^n=2^n\)

\(\Rightarrow\frac{1}{8}=2^n:16^n\)

\(\Rightarrow\frac{1}{8}=\left(\frac{2}{16}\right)^n\)

\(\Rightarrow\frac{1}{8}=\left(\frac{1}{8}\right)^n\)

\(\Rightarrow n=1\)

Vậy n=1

\(\frac{2^{4n}}{2^3}=2^n\Leftrightarrow2^{4n-3}=2^n\Rightarrow4n-3=n\Rightarrow n=1\)

\(3^3< 3^n< 3^5\Rightarrow n=4\)

\(\frac{1}{27}=3^{\frac{1}{81}}\)
=> \(n=\frac{1}{81}\)

\(\frac{16}{2^n}=\frac{1}{2}=\frac{16}{32}=\frac{16}{2^5}\)

=> n = 5

32 < 2ⁿ < 128

=> 2⁵ < 2ⁿ < 2⁷

=> 2ⁿ = 2⁶

=> n = 6