2005 AMC 12A Problem 23

Let $a = 2^j$ and $b = 2^k.$ Then $\log_a b = \dfrac{k}{j},$ which is an integer exactly when $k$ is a multiple of $j.$ For each $j,$ the number of valid $k \ne j$ in $\{1, \ldots, 25\}$ is $\left\lfloor \tfrac{25}{j} \right\rfloor - 1.$ Summing over $j$ gives $$\begin{aligned} &24 + 11 + 7 + 5 + 4 + 3 + 2 \\ &\quad {}+ 2 + 4 \cdot 1 = 62 \end{aligned}$$ ordered pairs $(a, b).$ Since there are $25 \cdot 24 = 600$ ordered pairs of distinct elements, the probability is $\dfrac{62}{600} = \dfrac{31}{300}.$ Thus, the correct answer is B.