smc

By the rules of logarithms, ${\log }_{10}(x^2+3)-2{\log }_{10}x={\log }_{10}(\frac{x^2+3}{x^2})={\log }_{10}(1+\frac{3}{x^2})$. As $x$ goes to infinity, $1+\frac{3}{x^2}$ gets arbitrarily close to $1$ (without ever reaching it), so ${\log }_{10}(1+\frac{3}{x^2})$ gets arbitrarily close to ${\log }_{10}(1)=0$ (without ever reaching it). Furthermore, because $1+\frac{3}{x^2}>1$, $\log (1+\frac{3}{x^2})$ is never negative. Thus, we can choose a real $x>\frac{2}{3}$ such that the given expression is $\boxed{\text{smaller than any positive number that might be specified}}$.