2024 AMC 12B 第 8 题

Divide numerator and denominator by $\log_2 x \cdot \log_3 x$ to turn the left side into $\dfrac{1}{\frac{1}{\log_2 x} + \frac{1}{\log_3 x}}$

Since $\dfrac{1}{\log_2 x} = \log_x 2$ and $\dfrac{1}{\log_3 x} = \log_x 3,$ the equation becomes $\log_x 6 = \dfrac12$

Dividing top and bottom by $\log_2 x \cdot \log_3 x,$ the left side becomes $$\begin{gathered} \frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x}} \\ = \frac{1}{\log_x 2 + \log_x 3} \\ = \frac{1}{\log_x 6}. \end{gathered}$$ So $\dfrac{1}{\log_x 6} = 2,$ meaning $\log_x 6 = \dfrac12,$ i.e. $x^{\frac{1}{2}} = 6.$

Therefore $x = 36.$

Thus, the correct answer is C.