AMC12 2002 A

AMC12 2002 A · Q14

AMC12 2002 A · Q14. It mainly tests Logarithms (rare).

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?

对所有正整数 $n$，定义 $f(n)=\log_{2002} n^2$。令 $N=f(11)+f(13)+f(14)$。下列哪个关系成立？

(A) N > 1 N > 1

(B) N = 1 N = 1

(D) N = 2 N = 2

(E) N > 2 N > 2

Answer

Correct choice: (D)

正确答案：（D）

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 $N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{(D) N=2}$.

首先注意到 $2002 = 11 \cdot 13 \cdot 14$。利用任意底数下都有 $\log a + \log b = \log ab$，可得 $N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{(D) N=2}$。

Topics

AL.015 Logarithms (rare)