Question

已知 $4^{x_1} = 5$ 、 $5^{x_2} = 6$ 、 $6^{x_3} = 7, \ldots, 127^{x_{124}} = 128$ 。求 $x_1 x_2 \cdots x_{124}$ 。

Suppose that $4^{x_1} = 5,$ $5^{x_2} = 6,$ $6^{x_3} = 7, \ldots, 127^{x_{124}} = 128.$ What is $x_1 x_2 \cdots x_{124}?$

$2$

$\dfrac{5}{2}$

$3$

$\dfrac{7}{2}$

$4$

由 $4^{x_1} = 5$ 得 $x_1 = \log_4 5$ ，一般地 $x_k = \log_{k+3}(k+4)$ 。

这个乘积会望远镜相消： $\begin{aligned} &x_1 x_2 \cdots x_{124} \\ &= \log_4 5 \cdot \log_5 6 \cdots \log_{127} 128 \\ &= \log_4 128. \end{aligned}$

因为 $128 = 2^7$ 且 $4 = 2^2$ ，结果为 $\dfrac{7\log 2}{2\log 2} = \dfrac72$ 。

所以正确答案是 D。

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.