8. $[\mathbf{6}]$ Let $A:=\mathbb{Q} \backslash\{0,1\}$ denote the set of all rationals other than 0 and 1. A function $f: A \rightarrow \mathbb{R}$ has the property that for all $x \in A$,
$$
f(x)+f\left(1-\frac{1}{x}\right)=\log |x| .
$$

Compute the value of $f(2007)$.
Answer: $\log (\mathbf{2 0 0 7} / \mathbf{2 0 0 6})$. Let $g: A \rightarrow A$ 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
$$
f(x)=\frac{1}{2} \cdot(\log |x|-\log |g(x)|+\log |g(g(x))|) .
$$

For $x=2007$, we have $g(x)=\frac{2006}{2007}$ and $g(g(x))=\frac{-1}{2006}$, so that
$$
f(2007)=\frac{\log |2007|-\log \left|\frac{2006}{2007}\right|+\log \left|\frac{-1}{2006}\right|}{2}=\log (2007 / 2006) .
$$