smc

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 $\boxed{6}$.