jmc

Let $g:A\to A$ be defined by $g(x):=1-1/x$; the key property is that $$g(g(g(x)))=1-\frac{1}{1-\frac{1}{1-\frac{1}{x}}}=x.$$The given equation rewrites as $f(x)+f(g(x))=\log |x|$. Substituting $x=g(y)$ and $x=g(g(z))$ gives the further equations $f(g(y))+f(g)g(y))=\log |g(x)|$ and $f(g)g(z))+f(z)=\log |g(g(x))|.$ Setting $y$ and $z$ to $x$ and solving the system of three equations for $f(x)$ gives $$f(x)=\frac{1}{2}\cdot \left(\log |x|-\log |g(x)|+\log |g(g(x))|\right).$$For $x=2007$, we have $g(x)=\frac{2006}{2007}$ and $g(g(x))=\frac{-1}{2006}$, so that $$f(2007)=\frac{\log |2007|-\log \left|\frac{2006}{2007}\right|+\log \left|\frac{-1}{2006}\right|}{2}=\boxed{\log \left(\frac{2007}{2006}\right)}.$$