2014 AMC 12B Problem 20

Below is the professionally curated solution for Problem 20 of the 2014 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12B solutions, or check the answer key.

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Concepts:logarithm inequality quadratic

Difficulty rating: 2110

20.

For how many positive integers $x$ is $\log_{10}(x - 40) + \log_{10}(60 - x) \lt 2?$

$10$

$18$

$19$

$20$

infinitely many

Solution:

The logarithms are defined only when $x - 40 \gt 0$ and $60 - x \gt 0,$ so $40 \lt x \lt 60.$

Within this range the inequality becomes $(x-40)(60-x) \lt 100,$ which expands to $x^2 - 100x + 2500 \gt 0,$ i.e. $(x-50)^2 \gt 0.$ This holds for every $x \ne 50.$

The integers strictly between $40$ and $60$ except $50$ are $41, \ldots, 49$ and $51, \ldots, 59,$ which is $18$ values.

Thus, the correct answer is B.

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