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2024 AMC 12B Problem 8

Problem 8 of 25EasierAlgebra

What value of xx satisfies log2xlog3xlog2x+log3x=2?\frac{\log_2 x \cdot \log_3 x}{\log_2 x + \log_3 x} = 2?

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Solution

Dividing top and bottom by log2xlog3x,\log_2 x \cdot \log_3 x, the left side becomes 11log2x+1log3x=1logx2+logx3=1logx6. \begin{gathered} \frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x}} \\ = \frac{1}{\log_x 2 + \log_x 3} \\ = \frac{1}{\log_x 6}. \end{gathered} So 1logx6=2,\dfrac{1}{\log_x 6} = 2, meaning logx6=12,\log_x 6 = \dfrac12, i.e. x12=6.x^{\frac{1}{2}} = 6. Therefore x=36.x = 36. Thus, the correct answer is C.

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Concepts: logarithm

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.