AMC 12 A

Solution:
Denote the edge lengths by $a = \log_2 x$, $b = \log_3 x$, and $c = \log_4 x$. The condition that the surface area numerically equals the volume is equivalent to $2(ab + ac + bc) = abc$. Dividing both sides by $2abc$ gives $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}$. By the Change of Base Formula, $\frac{1}{a} = \log_2 x$, $\frac{1}{b} = \log_3 x$, and $\frac{1}{c} = \log_4 x$. Therefore $\log_2 2 + \log_3 3 + \log_4 4 = \frac{1}{2}$, so $\log_2(2 \cdot 3 \cdot 4) = \frac{1}{2}$. From $x^{\frac{1}{2}} = 24$ it follows that $x = 24^2 = 576$.

Because the surface area and volume are numerically equal,
$$2 (\log_2 x \log_3 x + \log_2 x \log_4 x + \log_3 x \log_4 x) = \log_2 x \log_3 x \log_4 x.$$
By the Change of Base Formula with base 10,
$$2 \left( \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 3} + \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 4} + \frac{\log x}{\log 3} \cdot \frac{\log x}{\log 4} \right) = \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 3} \cdot \frac{\log x}{\log 4}.$$
Multiplying both sides by $\log 2 \cdot \log 3 \cdot \log 4$ and simplifying gives
$$2 (\log x)^2 (\log 4 + \log 3 + \log 2) = (\log x)^3.$$
Using log properties and moving all terms to one side gives
$$(\log x)^2 (2 \log (4 \cdot 3 \cdot 2) - \log x) = 0.$$
Thus either $\log x = 0$ or $2\log(4 \cdot 3 \cdot 2) - \log x = 0$. If $\log x = 0$ then $x = 1$, but this would make each edge length of the prism 0, which is impossible. Therefore $2\log(4 \cdot 3 \cdot 2) - \log x = 0$. This means $\log x = \log (24^2)$ and it follows that $x = 24^2 = 576$.