smc

We rewrite the logarithms in the problem. $$\log (x)+3\log (y)=1$$ $$2\log (x)+\log (y)=1$$ $$\log (x)+\log (y)=c$$ where $c$ is the desired quantity. Set $u=\log (x)$ and $v=\log (y)$. Then we have that $$u+3y=1\text{(1)}$$ $$2u+y=1\text{(2)}$$ $$u+v=c$$. Notice that $$2\cdot \text{(2)}+\text{(1)}=5u+5v=3⟹u+v=\frac{3}{5}⟹c=\boxed{\frac{3}{5}}$$.