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\(8^n:2^n=16^{2011}\)
\(\left(2^3\right)^n:2^n=\left(2^4\right)^{2011}\)
\(2^{3n}:2^n=2^{8044}\)
\(2^{3n-n}=2^{8044}\)
\(\Rightarrow3n-n=8044\)
\(2n=8044\)
\(\Rightarrow n=\frac{8044}{2}\)
\(n=4022\)
Vậy \(n=4022\)
\(\left(2^3\right)^n\)\(:2^n\)\(=\left(2^4\right)^{2021}\)
\(2^{3n}\)\(:2^n\)\(=2^{4x2021}\)\(=2^{8084}\)
\(2^{3n-n}\)\(=2^{8084}\)
\(=>3n-n=8084\)
\(2n=8084\)
\(n=8084:2=4042\)
\(=>n=4042\)
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: \(n=4\)
`1/8 xx 16^n =2^n`
`1/(2^3) xx (2^4)^n =2^n`
` 2^(-3) xx 2^(4n) =2^n`
` 2^(4n-3) =2^n`
`4n-3=n`
`3n=3`
`n=1`
\(a,\frac{16}{2^n}=2=>\frac{2^4}{2^n}=2=>2^4:2^n=2=>2^{4-n}=2=>4-n=1=>n=3\)
\(b,\frac{\left(-3\right)^n}{81}=-27=>\frac{\left(-3\right)^n}{3^4}=\left(-3\right)^3=>\frac{\left(-3\right)^n}{\left(-3\right)^4}=\left(-3\right)^3=>\left(-3\right)^{n-4}=\left(-3\right)^3=>n-4=3=>n=7\)
\(c,8^n:2^n=4=>\left(8:2\right)^n=4=>4^n=4=>n=1\)
a) Ta có: \(8^n:2^n=16^{2011}\)
\(\Leftrightarrow4^n=\left(4^2\right)^{2011}\)
\(\Leftrightarrow n=4022\)
b) Ta có: \(2^n+2^{n+3}=144\)
\(\Leftrightarrow2^n\left(1+2^3\right)=144\)
\(\Leftrightarrow2^n=16\)
hay n=4
\(8^n\div2^n=16^{2011}\)
\(\left(8\div2\right)^n=\left(4^2\right)^{2011}\)
\(4^n=4^{4022}\)
\(\Rightarrow n=4022\)
mình nghĩ ý b là
\(2^n+2^{n+3}=144\)
\(2^n+2^n\cdot2^3=144\)
\(2^n\left(1+8\right)=144\)
\(2^n\cdot9=144\)
\(2^n=16\)
\(2^n=2^4\)
\(\Rightarrow n=4\)