If $${4^{4x + 1}} = \frac{1}{{64}}{\text{,}}$$   then the value of x is?

$$\eqalign{ & {4^{4x + 1}} = \frac{1}{{64}} \cr & \Rightarrow {4^{4x + 1}} = \frac{1}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow {4^{4x + 1}} = {\left( 4 \right)^{ - 3}} \cr & \Rightarrow 4x + 1 = - 3 \cr & \Rightarrow 4x = - 4 \cr & \Rightarrow x = - 1 \cr} $$