jmc

Add the two equations to get that $\log x+\log y+2(\log (\gcd (x,y))+\log (\text{lcm}(x,y)))=630$. Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log (\gcd (x,y))+\log (\text{lcm}(x,y)))=630$. Using the theorem that $\gcd (x,y)\cdot \text{lcm}(x,y)=x\cdot y$, along with the previously mentioned theorem, we can get the equation $3\log (xy)=630$. This can easily be simplified to $\log (xy)=210$, or $xy=10^{210}$. $10^{210}$ can be factored into $2^{210}\cdot 5^{210}$, and $m+n$ equals to the sum of the exponents of $2$ and $5$, which is $210+210=420$. Multiply by two to get $2m+2n$, which is $840$. Then, use the first equation ($\log x+2\log (\gcd (x,y))=60$) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$ (you can also test when $x>y$, which is a contradiction to the restrains you set before). Therefore, $\gcd (x,y)=x$. Then, turn the equation into $3\log x=60$, which yields $\log x=20$, or $x=10^{20}$. Factor this into $2^{20}\cdot 5^{20}$, and add the two 20's, resulting in $m$, which is $40$. Add $m$ to $2m+2n$ (which is $840$) to get $40+840=\boxed{880}$.