jmc

Let $a=f(2)$ and $b=f(5).$ Setting $x=2$ and $y=5,$ we get $$14=f(10)=f(2)+f(5)=a+b.$$Setting $x=10$ and $y=2,$ we get $$f(20)=f(10)+f(2)=a+b+a=2a+b.$$Setting $x=20$ and $y=2,$ we get $$20=f(40)=f(20)+f(2)=2a+b+a=3a+b.$$Solving the system $a+b=14$ and $3a+b=20,$ we find $a=3$ and $b=11.$ Then $$\begin{array}{rl}f(500)& =f(2\cdot 2\cdot 5\cdot 5\cdot 5)\\ & =f(2)+f(2)+f(5)+f(5)+f(5)\\ & =2\cdot 3+3\cdot 11\\ & =\boxed{39}.\end{array}$$