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

           \(\frac{1}{2^3}.\left(2^4\right)^n=2^n\)

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

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

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

                       \(\frac{1}{2^3}=2^{n\left(1-4\right)}\)

                       \(\frac{1}{2^3}=2^{\left(-3\right)n}\)

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

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

          \(2^{3\left(-n+1\right)}=2^0\)

\(\Rightarrow3\left(-n+1\right)=0\)

      \(\Rightarrow-n+1=0\)

                      \(-n=-1\)

                          \(n=1\)

\(\frac{1}{8}\cdot16^n=2^n\)

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

=> \(\frac{2^{4n}}{2^3}=2^n\)

=> \(2^{4n-3}=2^n\)

=> \(4n-3=n\)

=> \(n=1\)

Vậy n = 1

\(8.16\ge2^n\ge4\)  => \(2^3.2^4\ge2^n\ge2^2\)=> \(2^7\ge2^n\ge2^2\)

=> \(7\ge n\ge2\)

=> \(n\in\left\{2;3;4;5;6;7\right\}\)

\(8.16\ge2^n\ge4\)

\(\Leftrightarrow2^3.2^4\ge2^n\ge2^2\)

\(\Leftrightarrow2^7\ge2^n\ge2^2\)

\(\Rightarrow2\le n\le7\)

\(\Rightarrow n\varepsilon\left\{2;3;4;5;6;7\right\}\)

\(8\times16\ge2^n\ge4\)

\(\Leftrightarrow128\ge2^n\ge4\)

\(\Leftrightarrow2^7\ge2^n\ge2^2\)

\(\Leftrightarrow7\ge x\ge2\)

\(\Leftrightarrow x\in(7,6,5,4,3,2)\)

Vậy \(x\in(7,6,5,4,3,2)\)

A, 1/8=2^x:16^x

2^-3=2^(x-4x)

x-4x=-3

x*(1-4)=-3

x*(-3)=-3

x=1

a, \(\frac{1}{8}.16^x=2^x\)

    \(\frac{1}{8}=\frac{2^x}{16^x}\)

     \(\frac{1}{8}=\frac{1^x}{8^x}\)

=> 8 = 8^x

=> x = 1