smc

Notice that ${\log }_{5}10={\log }_{5}2+1=\frac{1}{{\log }_{2}5}+1$. So we are trying to find ${\log }_{2}5$. Denote ${\log }_{10}2$ as $x$. Then $\frac{1}{x}={\log }_{2}10={\log }_{2}5+1$. Therefore ${\log }_{2}5=\frac{1}{x}-1$, and plugging this in gives ${\log }_{5}10=\frac{1}{\frac{1}{x}-1}+1=\frac{x}{1-x}+1=\frac{1}{1-x}$. Since $x$ is around $\frac{3}{10}$, we substitute and get $\boxed{\frac{10}{7}}$.