Expressing $\log(\frac{x}{x+y})$ using known value of $\frac{\log(x)}{\log(x) + \log(y)}$

Question

I have the quantity $\frac{\log(x)}{\log(x) + \log(y)}$ What I need is to calculate $\log(\frac{x}{x+y})$ Is it possible?

Any further help would be much appreciated!

Answer 1

$\frac{\log(x)}{\log(x)+\log(y)}$ is invariant under the map $(x,y)\to (x^2,y^2)$ while $\log\frac{x}{x+y}$ is not, hence in general you cannot compute $\log\frac{x}{x+y}$ from $\frac{\log(x)}{\log(x)+\log(y)}$.

Answer 2

$$ \frac{\log(2)}{\log(2)+\log(4)}=\frac13\quad\text{and}\quad\frac{\log(3)}{\log(3)+\log(9)}=\frac13 $$ but $$ \log\left(\frac{2}{2+4}\right)=\log\left(\frac13\right)\ne\log\left(\frac14\right)=\log\left(\frac{3}{3+9}\right) $$ so $\log\left(\frac{x}{x+y}\right)$ cannot be a function of $\frac{\log(x)}{\log(x)+\log(y)}$