smc

Take the logarithm base $x$ of both sides to get ${\log }_{10}x={\log }_{x}10$. Use the change of base formula to get ${\log }_{10}x=\frac{{\log }_{10}10}{{\log }_{10}x}=\frac{1}{{\log }_{10}x}$. Therefore, $({\log }_{10}x{)}^2=1$ and ${\log }_{10}x=\pm 1$. From ${\log }_{10}x=1$, we have $x=10^1=10$ and from ${\log }_{10}x=-1$ we have $x=10^{-1}=\frac{1}{10}$. The product is thus $10\times \frac{1}{10}=1,\boxed{1}$.