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Difference between revisions of "2018 AMC 12A Problems/Problem 14"

Problem

The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$, can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$? $\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35$

Solution 1

Base switch to log 2 and you have $\frac{\log_2 4}{\log_2 3x} = \frac{\log_2 8}{\log_2 2x}$ . $\frac{2}{\log_2 3x} = \frac{3}{\log_2 2x}$ $2*\log_2 2x = 3*\log_2 3x$

Then $\log_2 (2x)^2 = \log_2 (3x)^3$. so $4x^2=27x^3$ and we have $x=\frac{4}{27}$ leading to $\boxed{ (D) 31}$ (jeremylu)

Solution 2

If you multiply both sides by $\log_2 (3x)$

then it should come out to $\log_2 (3x)$ * $\log_{3x} (4)$ = $\log_2 {3x}$ * $\log_{2x} (8)$

that then becomes $\log_2 (4)$ * $\log_{3x} (3x)$ = $\log_2 (8)$ * $\log_{2x} (3x)$

which simplifies to 2*1 = 3 $\log_{2x} (3x)$

so now $\frac{2}{3}$ = $\log_{2x} (3x)$ putting in exponent form gets $(2x)^2$ = $(3x)^3$

so 4 $x^2$ = 27 $x^3$

dividing yields x = 4/27 and

4+27 = $\boxed{ (D) 31}$

- Pikachu13307

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