2014 AMC 12B Problem 20

Attempt Problem 20 of the 2014 AMC 12B 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 2014 AMC 12B solutions, or check the answer key.

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

20.

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

$10$

$18$

$19$

$20$

infinitely many

Answer: B

Concepts:logarithm inequality quadratic

Difficulty rating: 2110

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.

← Problem 19#19

Full Exam

Problem 21#21 →

Problem 20 in Other Years

1999 AMC 12 · 2000 AMC 12 · 2001 AMC 12 · 2002 AMC 12A · 2002 AMC 12B · 2003 AMC 12A · 2003 AMC 12B · 2004 AMC 12A · 2004 AMC 12B · 2005 AMC 12A · 2005 AMC 12B · 2006 AMC 12A · 2006 AMC 12B · 2007 AMC 12A · 2007 AMC 12B · 2008 AMC 12A · 2008 AMC 12B · 2009 AMC 12A · 2009 AMC 12B · 2010 AMC 12A · 2010 AMC 12B · 2011 AMC 12A · 2011 AMC 12B · 2012 AMC 12A · 2012 AMC 12B · 2013 AMC 12A · 2013 AMC 12B · 2014 AMC 12A · 2015 AMC 12A · 2015 AMC 12B · 2016 AMC 12A · 2016 AMC 12B · 2017 AMC 12A · 2017 AMC 12B · 2018 AMC 12A · 2018 AMC 12B · 2019 AMC 12A · 2019 AMC 12B · 2020 AMC 12A · 2020 AMC 12B · 2021 AMC 12A Spring · 2021 AMC 12B Spring · 2021 AMC 12A Fall · 2021 AMC 12B Fall · 2022 AMC 12A · 2022 AMC 12B · 2023 AMC 12A · 2023 AMC 12B · 2024 AMC 12A · 2024 AMC 12B · 2025 AMC 12A · 2025 AMC 12B