Question

$2^{4x+1} . 8^{-x+3} = \frac{1}{16}$

Answer 1

$2^{4x+1}.8^{-x+3}=\frac{1}{16}\\ \\ 2^{4x+1}.2^{-3x+9}=2^{-4}\\ \\ 2^{4x+1-3x+9}=2^{-4}\\ \\ 4x+1-3x+9=-4\\ \\ \boxed{x=-15}$