Since the line $x=k$ is vertical, we are only concerned with vertical distance.
In other words, we want to find the value of $k$ for which the distance $|\log_5 x - \log_5 (x+4)| = \frac{1}{2}$
Since $\log_5 x$ is a strictly increasing function, we have:
$\log_5 (x + 4) - \log_5 x = \frac{1}{2}$
$\log_5 (\frac{x+4}{x}) = \frac{1}{2}$
$\frac{x+4}{x} = 5^\frac{1}{2}$
$x + 4 = x\sqrt{5}$
$x\sqrt{5} - x = 4$
$x = \frac{4}{\sqrt{5} - 1}$
$x = \frac{4(\sqrt{5} + 1)}{5 - 1^2}$
$x = 1 + \sqrt{5}$
The desired quantity is $1 + 5 = 6$, and the answer is $\boxed{A}$.