2002 AMC 12A Problem 14

Attempt Problem 14 of the 2002 AMC 12A below, then check your answer against the professionally curated solution 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 problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

For all positive integers n,n, let f(n)=log⁡2002n2.f(n) = \log_{2002} n^2. Let N=f(11)+f(13)+f(14).N = f(11) + f(13) + f(14). Which of the following relations is true?
N>1N \gt 1
N=1N = 1
1<N<21 \lt N \lt 2
N=2N = 2
N>2N \gt 2
Answer: D
Concepts:logarithm
Small hint:
f(n)=log⁡2002n2=2log⁡2002nf(n) = \log_{2002} n^2 = 2\log_{2002} n
Big hint:
Combine into N=log⁡2002(11⋅13⋅14)2N = \log_{2002}(11\cdot 13\cdot 14)^2 and evaluate the product inside
Solution:
Using log⁡a2=2log⁡a\log a^2 = 2\log a and adding logs, N=log⁡2002112+log⁡2002132+log⁡2002142=log⁡2002(11⋅13⋅14)2. \begin{aligned} N &= \log_{2002} 11^2 + \log_{2002} 13^2 \\ &\quad {}+ \log_{2002} 14^2 \\ &= \log_{2002}(11\cdot 13\cdot 14)^2. \end{aligned}
Since 11⋅13⋅14=2002,11\cdot 13\cdot 14 = 2002, this is log⁡200220022=2.\log_{2002} 2002^2 = 2.
Thus, the correct answer is D.
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