The solution to the equation log_3x 4 = log_2x 8, where x is a positive real number other than 13 or 12, can be written as (p)/(q), where p and q are relatively prime positive integers. What is p + q?

2018 AMC 12A Problem 14

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14.

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?

5

13

17

31

35

Rewrite both sides with a common base:

\log_{3x} 4 = \tfrac{\log 4}{\log 3x},

and similarly for

\log_{2x} 8

Cross-multiplying reduces the equation to

(2x)^2 = (3x)^3;

solve for

x

as a reduced fraction

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.