2024 AMC 12B Problems/Problem 8
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 |
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All AMC 12 Problems and Solutions |
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