jmc

In order to find ${\log }_{11}x$, we must first find the value of $x$. We begin by writing ${\log }_5(x+4)=3$ in exponential form, which gives us $5^3=x+4$. Solving for $x$, we find that $x=5^3-4=125-4=121$. After plugging this value of $x$ into the second expression, we now need to find ${\log }_{11}121$. Since we know that $11^2=121$, ${\log }_{11}121=\boxed{2}$.