2018 AMC 12A Problem 14

Problem 14 of 25IntermediateAlgebra

The solution to the equation

\log_{3x} 4 = \log_{2x} 8,

where

x

is a positive real number other than

\tfrac13

\tfrac12,

can be written as

\tfrac{p}{q},

where

p

and

q

are relatively prime positive integers. What is

p + q?

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Solution

Writing both logarithms in base

2:

\tfrac{2}{\log_2 3x} = \tfrac{3}{\log_2 2x},

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.