Difference between revisions of "2002 AMC 12A Problems/Problem 14"

Problem

For all positive integers $n$, let $f(n)=\log_{2002} n^2$. Let $N=f(11)+f(13)+f(14)$. Which of the following relations is true?

$\text{(A) }N<1 \qquad \text{(B) }N=1 \qquad \text{(C) }12$

Solution

First, note that $2002 = 11 \cdot 13 \cdot 14$.

Using the fact that for any base we have $\log a + \log b = \log ab$, we get that $N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{(D) 2}$.