1967 AMC 12 Problem 23

Attempt Problem 23 of the 1967 AMC 12 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 1967 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

If xx is real and positive and grows beyond all bounds, then log⁡3(6x−5)−log⁡3(2x+1)\log_3(6x-5)-\log_3(2x+1) approaches:
00
11
33
44
no finite number
Answer: B
Concepts:logarithmbounding to limit cases
Small hint:
Combine the logarithms into one logarithm
Big hint:
Find the limit of 6x−52x+1\frac{6x-5}{2x+1}
Solution:
The difference is log⁡3(6x−52x+1). \log_3\left(\frac{6x-5}{2x+1}\right). The fraction approaches 3,3, so the expression approaches log⁡33=1.\log_3 3=1.
Therefore, the correct answer is B.
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