2005 AMC 12B Problem 13

From $4^{x_1} = 5$ we get $x_1 = \log_4 5,$ and in general $x_k = \log_{k+3}(k+4).$ The product telescopes: $$\begin{aligned} &x_1 x_2 \cdots x_{124} \\ &= \log_4 5 \cdot \log_5 6 \cdots \log_{127} 128 \\ &= \log_4 128. \end{aligned}$$ Since $128 = 2^7$ and $4 = 2^2,$ this equals $\dfrac{7\log 2}{2\log 2} = \dfrac72.$ Thus, the correct answer is D.