2024 AMC 12A Problems/Problem 15
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Problem
Suppose 
and 
are real numbers for which ![\[\log_{p}(q) - \log_{2p}(q) =\frac{1}{3} \qquad \operatorname{and} \qquad \log_{2p}(q) - \log_{4p}(q) =\frac{1}{4}.\]](http://latex.artofproblemsolving.com/8/1/3/813cdbfad6dc9278751c7c8e487b83ddc95eb3f5.png)
What is the value of 
?
Solution
Let 
and 
. Then note ![\[\log_{p}(q) - \log_{2p}(q) =\frac{\log_{2}(q)}{\log_{2}(p)} -\frac{\log_{2}q}{\log_{2}(2p)} =\frac{y}{x} -\frac{y}{x + 1} =\frac{y}{x(x + 1)} =\frac{1}{3} \implies x(x + 1) = 3y.\]](http://latex.artofproblemsolving.com/b/1/6/b16fd641df70c288d26c9674aba5d2994627f0e6.png)
Similarly, ![\[\log_{2p}(q) - \log_{4p}(q) =\frac{\log_{2}(q)}{\log_{2}(2p)} -\frac{\log_{2}(q)}{\log_{2}(4p)} =\frac{y}{x + 1} -\frac{y}{x + 2} =\frac{y}{(x + 1)(x + 2)} =\frac{1}{4} \implies (x + 1)(x + 2) = 4y.\]](http://latex.artofproblemsolving.com/0/f/0/0f027b5c42205e60a9f4c57b05758709fe0af7c0.png)
Dividing the two equations, 
which solves to 
. Substituting, 
and ![\[\log_{4p}(q) - \log_{8p}(q) =\frac{y}{x + 2} -\frac{y}{x + 3} = \frac{14}{8} - \frac{14}{9} = \boxed{\textbf{(D)}~\frac{7}{36}}.\]](http://latex.artofproblemsolving.com/3/8/3/38333aa04376a219fa60a3baab678d30b0dfbe31.png)
See also
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
