Question

方程组的解为 $(x_1, y_1)$ 和 $(x_2, y_2)$ 。求 $\log_{30}\left(x_1 y_1 x_2 y_2\right)$ 。 $\log_{225} x + \log_{64} y = 4$ $\log_x 225 - \log_y 64 = 1$

The solutions to the system of equations $\log_{225} x + \log_{64} y = 4$ $\log_x 225 - \log_y 64 = 1$ are $(x_1, y_1)$ and $(x_2, y_2).$ Find $\log_{30}\left(x_1 y_1 x_2 y_2\right).$

答案

答案：12

知识点：对数韦达定理换元法

难度评级：2360

解答：

设 $p = \log_{225} x$ 、 $q = \log_{64} y$ ，则 $\log_x 225 = \frac{1}{p}$ 且 $\log_y 64 = \frac{1}{q}$ 。方程组变为 $p + q = 4$ 和 $\frac{1}{p} - \frac{1}{q} = 1$ 。将 $q = 4 - p$ 代入第二个方程并清分母，得到 $4 - 2p = p(4 - p)$ ，即 $p^2 - 6p + 4 = 0$ 。

原方程组的两个解对应这个二次方程的两个根，所以由韦达定理 $p_1 + p_2 = 6$ ，进而 $q_1 + q_2 = 8 - 6 = 2$ 。于是所以 $\log_{30}\left(x_1 y_1 x_2 y_2\right) = 12$ 。 $\begin{aligned} x_1 y_1 x_2 y_2 &= 225^{p_1 + p_2} \cdot 64^{q_1 + q_2} \\ &= 225^6 \cdot 64^2 \\ &= 15^{12} \cdot 2^{12} \\ &= 30^{12}, \end{aligned}$

Let $p = \log_{225} x$ and $q = \log_{64} y,$ so $\log_x 225 = \frac{1}{p}$ and $\log_y 64 = \frac{1}{q}.$ The system becomes $p + q = 4$ and $\frac{1}{p} - \frac{1}{q} = 1.$ Substituting $q = 4 - p$ into the second equation and clearing denominators gives $4 - 2p = p(4 - p),$ that is, $p^2 - 6p + 4 = 0.$

The two solutions of the system correspond to the two roots of this quadratic, so by Vieta's formulas $p_1 + p_2 = 6,$ and then $q_1 + q_2 = 8 - 6 = 2.$ Hence $\begin{aligned} x_1 y_1 x_2 y_2 &= 225^{p_1 + p_2} \cdot 64^{q_1 + q_2} \\ &= 225^6 \cdot 64^2 \\ &= 15^{12} \cdot 2^{12} \\ &= 30^{12}, \end{aligned}$ so $\log_{30}\left(x_1 y_1 x_2 y_2\right) = 12.$

2002 AIME I 第 6 题

答案：12

难度评级：2360

解答：

其他年份的第 6 题