AMC 12 B, 2024

Answer (C): Observe that
$$\begin{aligned} 2 &= \frac{\log_2 x \cdot \log_3 x}{\log_2 x + \log_3 x} \\ &= \frac{1}{\frac{1}{\log_3 x} + \frac{1}{\log_2 x}} \\ &= \frac{1}{\log_4 3 + \log_4 2} \\ &= \frac{1}{\log_4 6} = \log_6 x. \end{aligned}$$
It follows that $x = 6^2 = 36$.

The given equation is equivalent to
$$\log_2 x \cdot \log_3 x = 2 \log_2 x + 2 \log_3 x.$$
Note that $\log_3 x = \frac{\log_2 x}{\log_2 3}$, so
$$\log_2 x \cdot \frac{\log_2 x}{\log_2 3} = 2 \log_2 x + 2 \frac{\log_2 x}{\log_2 3}.$$
Multiplying both sides by $\frac{\log_2 3}{\log_2 x}$ gives
$$\log_2 x = 2 \log_2 3 + 2 = \log_2 9 + 2.$$
Then
$$x = 2^{\log_2 9+2} = 2^{\log_2 9} \cdot 2^2 = 9 \cdot 4 = 36.$$