jmc

By the change-of-base formula, $${\log }_{\sqrt{5}}(x^3-2)=\frac{{\log }_5(x^3-2)}{{\log }_5\sqrt{5}}=\frac{{\log }_5(x^3-2)}{1/2}=2{\log }_5(x^3-2),$$and $${\log }_{\frac{1}{5}}(x-2)=\frac{{\log }_5(x-2)}{{\log }_5\frac{1}{5}}=-{\log }_5(x-2),$$so the given equation becomes $$2{\log }_5(x^3-2)=4.$$Then ${\log }_5(x^3-2)=2,$ so $x^3-2=5^2=25.$ Then $x^3=27,$ so $x=\boxed{3}.$