1. Determine which of the numbers is greater:

$$
\log _{2010} 2011 \text { or } \log _{2011} 2012 .
$$
# Problem 1.

Answer: the first number is greater (in all cases).

Solution. Let's compare the numbers $\log _{k-1} k$ and $\log _{k}(k+1)$ for all $k \geqslant 3$. Note that by the Cauchy inequality: $\sqrt{\log _{k}(k-1) \cdot \log _{k}(k+1)}<0.5 \log _{k}\left(k^{2}-1\right)<1$. Therefore, $\log _{k}(k+1)<\log _{k-1} k$.