jmc

First we solve for $x$. Converting our logarithm to exponential form gives $25^{\frac{1}{2}}=x-4$. We know $25^{\frac{1}{2}}=\sqrt{25}=5$, so we have $5=x-4$ or $x=9$. Then we must find $\frac{1}{{\log }_{x}3}$ where $x=9$. Let ${\log }_{9}3=a$. Then $9^a=3$. Since $9=3^2$ we have $3^{2a}=3^1$ so $2a=1$ or $a=\frac{1}{2}$. We want to find $\frac{1}{{\log }_{9}3}=\frac{1}{a}=\frac{1}{\frac{1}{2}}=\boxed{2}$.