2024 AMC 12A Problems/Problem 15

Problem

Suppose $p$ and $q$ are real numbers for which $\log_{p}(q) - \log_{2p}(q) =\frac{1}{3} \qquad \operatorname{and} \qquad \log_{2p}(q) - \log_{4p}(q) =\frac{1}{4}.$ What is the value of $\log_{4p}(q) - \log_{8p}(q)$ ?

$\textbf{(A)}~\frac{1}{6}\qquad\textbf{(B)}~\frac{7}{40}\qquad\textbf{(C)}~\frac{8}{45}\qquad\textbf{(D)}~\frac{7}{36}\qquad\textbf{(E)}~\frac{1}{5}$

Solution

Let $x = \log_{2}p$ and $y = \log_{2}q$ . Then note $\log_{p}(q) - \log_{2p}(q) =\frac{\log_{2}(q)}{\log_{2}(p)} -\frac{\log_{2}q}{\log_{2}(2p)} =\frac{y}{x} -\frac{y}{x + 1} =\frac{y}{x(x + 1)} =\frac{1}{3} \implies x(x + 1) = 3y.$ Similarly, $\log_{2p}(q) - \log_{4p}(q) =\frac{\log_{2}(q)}{\log_{2}(2p)} -\frac{\log_{2}(q)}{\log_{2}(4p)} =\frac{y}{x + 1} -\frac{y}{x + 2} =\frac{y}{(x + 1)(x + 2)} =\frac{1}{4} \implies (x + 1)(x + 2) = 4y.$ Dividing the two equations, $\tfrac{x}{x + 2} = \tfrac{3}{4}$ which solves to $x = 6$ . Substituting, $y = \tfrac{1}{3} \cdot 6 \cdot 7 = 14$ and $\log_{4p}(q) - \log_{8p}(q) =\frac{y}{x + 2} -\frac{y}{x + 3} = \frac{14}{8} - \frac{14}{9} = \boxed{\textbf{(D)}~\frac{7}{36}}.$

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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2024 AMC 12A Problems/Problem 15

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