jmc

Let $y={\log }_{2}2018.$ Then by the change of base formula, for any $b,$ $${\log }_{2^b}2018=\frac{{\log }_{2}2018}{{\log }_2{2}^b}=\frac{{\log }_{2}2018}{b}.$$Thus, ${\log }_{4}2018=\frac{{\log }_{2}2018}{2}=\frac{y}{2},$ ${\log }_{8}2018=\frac{{\log }_{2}2018}{3}=\frac{y}{3},$ and ${\log }_{64}2018=\frac{{\log }_{2}2018}{6}=\frac{y}{6}.$ Thus, the left-hand side becomes $$y+\frac{y}{2}+\frac{y}{3}+\frac{y}{6}=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)y=2y.$$Taking $b=\frac{1}{2},$ we see that $${\log }_{2^{1/2}}2018=\frac{{\log }_{2}2018}{1/2}=2y,$$so the left-hand side equals ${\log }_{2^{1/2}}2018,$ or ${\log }_{\sqrt{2}}2018.$ Thus, $x=\boxed{\sqrt{2}}.$