Difference between revisions of "2024 AMC 12B Problems/Problem 8"
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\end{align*} | \end{align*} | ||
so <math>\boxed{\textbf{(C) }36}</math> | so <math>\boxed{\textbf{(C) }36}</math> | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC12 box|year=2024|ab=B|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Revision as of 01:42, 14 November 2024
Problem
What value of satisfies
Solution 1
We have
\begin{align*}
&\log_2x\cdot\log_3x=2(\log_2x+\log_3x) \\
&1=\frac{2(\log_2x+\log_3x)}{\log_2x\cdot\log_3x} \\
&1=2(\frac{1}{\log_3x}+\frac{1}{\log_2x}) \\
&1=2(\log_x3+\log_x2) \\
&\log_x6=\frac{1}{2} \\
&x^{\frac{1}{2}}=6 \\
&x=36
\end{align*}
so
See also
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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.