jmc

Because the point for which we are looking is on the $x$-axis, we know that it is of the form $(x,0)$. We apply the distance formula. The distance from A is $$\begin{array}{rl}\sqrt{(-2-x{)}^2+(0-0{)}^2}& =\sqrt{x^2+4x+4}\end{array}$$ The distance from B is $$\begin{array}{rl}\sqrt{(0-x{)}^2+(4-0{)}^2}& =\sqrt{x^2+16}\end{array}$$ Because the point is equidistant from A and B, we set the two distances equal: $x^2+4x+4=x^2+16$. Simplifying gives us $4x=12$, or $x=\boxed{3}$.