\(2^n.16-2^{n+1}=2^6-2^3\\ \Leftrightarrow2^n.2^4-2^n.2=2^6-2^3\\ \Leftrightarrow2^n\left(2^4-2\right)=2^6-2^3\\ \Leftrightarrow2^n=\dfrac{2^6-2^3}{2^4-2}\\ \Leftrightarrow2^n=\dfrac{2\left(2^5-2^2\right)}{2\left(2^3-1\right)}\\ \Leftrightarrow2^n=\dfrac{28}{6}\\ \Leftrightarrow2^n=4\\ \Leftrightarrow2^n=2^2\\ \Leftrightarrow n=2\)
\(2^n.16-2^{n+1}=2^6-2^3\\ \Leftrightarrow2^n.2^4-2^{n+1}=64-8\\ \Leftrightarrow2^{n+4}-2^{n+1}=56\\ \Leftrightarrow2^{n+1}.\left(2^3-1\right)=56\\ \Leftrightarrow2^{n+1}.7=56\\ \Leftrightarrow2^{n+1}=\dfrac{56}{7}=8=2^3\\ \Leftrightarrow n+1=3\\ \Leftrightarrow n=2\)
\(2^n\cdot16-2^{n+1}=2^6-2^3\\ 2^n\cdot16-2^n\cdot2= 2^3\cdot\left(8-1\right)\\ 2^n\cdot\left(16-2\right)=8\cdot7\\ 2^n\cdot14=8\cdot7\\ 2^n\cdot\left(14:7\right)=8\\ 2^n\cdot2=8\\ 2^{n+1}=2^3\\ =>n+1=3\\ n=3-1\\ n=2\)
