jmc

We can write $$\begin{array}{rl}2{\log }_{10}x-{\log }_x\frac{1}{100}& =2{\log }_{10}x+{\log }_{x}100\\ & =2{\log }_{10}x+{\log }_{x}10^2\\ & =2{\log }_{10}x+2{\log }_{x}10\\ & =2({\log }_{10}x+{\log }_{x}10)\\ & =2\left({\log }_{10}x+\frac{1}{{\log }_{10}x}\right).\end{array}$$By AM-GM, $${\log }_{10}x+\frac{1}{{\log }_{10}x}\ge 2,$$so $2\left({\log }_{10}x+\frac{1}{{\log }_{10}x}\right)\ge 4.$

Equality occurs when $x=10,$ so the minimum value is $\boxed{4}.$