jmc

Using the property ${\log }_a{b}^x=x{\log }_{a}b,$ we have $$\begin{array}{rl}\frac{2}{{\log }_{4}2000^6}+\frac{3}{{\log }_{5}2000^6}& =\frac{2}{6{\log }_{4}2000}+\frac{3}{6{\log }_{5}2000}\\ & =\frac{1}{3{\log }_{4}2000}+\frac{1}{2{\log }_{5}2000}.\end{array}$$Since ${\log }_{a}b=\frac{1}{{\log }_{b}a}$, we can then write $$\frac{1}{3{\log }_{4}2000}+\frac{1}{2{\log }_{5}2000}=\frac{1}{3}{\log }_{2000}4+\frac{1}{2}{\log }_{2000}5,$$which equals $${\log }_{2000}(4^{1/3}{5}^{1/2})={\log }_{2000}(2^{2/3}{5}^{1/2}).$$Since $2000=2^4{5}^3={\left(2^{2/3}{5}^{1/2}\right)}^6$, the expression $\boxed{\frac{1}{6}}$.