Question

\(2^{x+1}=\dfrac{64.16}{2^8}\)

Answer 1

\(2^{x+1}=\dfrac{64.16}{2^8}\\ \Rightarrow2^{x+1}=\dfrac{2^6.2^4}{2^8}\\ \Rightarrow2^{x+1}=2^2\\ \Rightarrow x+1=2\\ \Rightarrow x=1\)