Lauren sent in her solution:

Since ${\log }_{b}x={\log }_b(y^{{\log }_{y}x})={\log }_{y}x{\log }_{b}y$, we have ${\log }_{y}x=\frac{{\log }_{b}x}{{\log }_{b}y}$ and (letting $b=x$) $1={\log }_{y}x{\log }_{x}y$.
Now ${\log }_{5}49=\frac{{\log }_{7}49}{{\log }_{7}5}=\frac{2}{{\log }_{7}5}$ and ${\log }_{7}125=\frac{{\log }_{5}125}{{\log }_{5}7}=\frac{3}{{\log }_{5}7}$, so
${\log }_{5}49\times {\log }_{7}125=\frac{2}{{\log }_{7}5}\times \frac{3}{{\log }_{5}7}=6$.