jmc

We use the distance formula: $$\sqrt{(4-1{)}^2+(x-3{)}^2}=\sqrt{3^2+(x-3{)}^2}=\sqrt{x^2-6x+18}=5.$$ Squaring both sides and rearranging terms, we find that $$\begin{array}{rl}{x}^2-6x+18& =25\\ x^2-6x-7& =0\\ (x-7)(x+1)& =0\end{array}$$ Thus, $x=7$ or $x=-1$. We are given that $x>0$, so $x=\boxed{7}$.

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Note that the points $(1,3)$, $(4,3)$, and $(4,x)$ form a right triangle. We are given that one of the legs has length 3 and the hypotenuse has length 5. This is a Pythagorean triple, so the last leg must have length 4. Therefore, $x=3-4=-1$ or $x=3+4=7$. Since $x>0$, we find that $x=\boxed{7}$.