8. [6] Which is greater, $\log _{2008}(2009)$ or $\log _{2009}(2010)$ ?
Answer: $\log _{2008} 2009$.
Solution: Let $f(x)=\log _{x}(x+1)$. Then $f^{\prime}(x)=\frac{x \ln x-(x+1) \ln (x+1)}{x(x+1) \ln ^{2} x}1$, so $f$ is decreasing. Thus $\log _{2008}(2009)$ is greater.