jmc

We can have ${\log }_{10}100=2$ and ${\log }_{10}1000=3$. Since ${\log }_{10}x$ increases as $x$ increases, we know that ${\log }_{10}100<{\log }_{10}579<{\log }_{10}1000$, meaning $2<{\log }_{10}579<3$. Thus, the desired sum is $2+3=\boxed{5}$.