2018 AMC 12A Problem 14

Below is the professionally curated solution for Problem 14 of the 2018 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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Concepts:logarithm

Difficulty rating: 1730

14.

The solution to the equation $\log_{3x} 4 = \log_{2x} 8,$ where $x$ is a positive real number other than $\tfrac13$ or $\tfrac12,$ can be written as $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p + q?$

$5$

$13$

$17$

$31$

$35$

Solution:

Writing both logarithms in base $2:$ $\tfrac{2}{\log_2 3x} = \tfrac{3}{\log_2 2x},$ so $2 \log_2 2x = 3 \log_2 3x,$ i.e. $(2x)^2 = (3x)^3.$ Then $4x^2 = 27x^3,$ giving $x = \tfrac{4}{27}.$ Since $\gcd(4, 27) = 1,$ we get $p + q = 4 + 27 = 31.$

Thus, the correct answer is D.

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