2002 AMC 12A Problem 14 — Solution | LIVE by Po-Shen Loh

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2002 AMC 12A Problem 14

Below is the professionally curated solution for Problem 14 of the 2002 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:logarithm

Difficulty rating: 1500

14.

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?

$N \gt 1$

$N = 1$

$1 \lt N \lt 2$

$N = 2$

$N \gt 2$

Solution:

Using $\log a^2 = 2\log a$ and adding logs, $N = \log_{2002} 11^2 + \log_{2002} 13^2 + \log_{2002} 14^2 = \log_{2002}(11\cdot 13\cdot 14)^2.$

Since $11\cdot 13\cdot 14 = 2002,$ this is $\log_{2002} 2002^2 = 2.$

Thus, the correct answer is D.

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