Question

设 $x$ 、 $y$ 、 $z$ 为满足 $\begin{aligned} 2\log_{x}(2y) &= 2\log_{2x}(4z) \\ &= \log_{2x^4}(8yz) \ne 0 \end{aligned}$ 的正实数。 $xy^5z$ 的值可表示为 $\frac{1}{2^{\frac{p}{q}}}$ ，其中 $p$ 和 $q$ 是互质正整数。求 $p + q$ 。

Let $x,$ $y,$ and $z$ be positive real numbers that satisfy $\begin{aligned} 2\log_{x}(2y) &= 2\log_{2x}(4z) \\ &= \log_{2x^4}(8yz) \ne 0. \end{aligned}$ The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{\frac{p}{q}}},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：49

知识点：对数指数换元法

难度评级：2650

小提示：

令 $x = 2^a$ 、 $y = 2^b$ 、 $z = 2^c$ ，这样每个对数都会变成关于指数的线性式之比

Set $x = 2^a,$ $y = 2^b,$ $z = 2^c$ so each logarithm becomes a ratio of linear expressions in the exponents

大提示：

两个比值相等时，它们也等于分子之和与分母之和的比值： $\frac{u_1}{v_1} = \frac{u_2}{v_2} = \frac{u_1 + u_2}{v_1 + v_2}$ 。把这个与第三个比值比较即可确定 $a$ 。

Equal ratios also equal their mediant: $\frac{u_1}{v_1} = \frac{u_2}{v_2} = \frac{u_1 + u_2}{v_1 + v_2}.$ Compare that with the third ratio to pin down $a.$

解答：

写成 $x = 2^a$ 、 $y = 2^b$ 、 $z = 2^c$ 。于是 $\log_x(2y) = \frac{b + 1}{a}$ ， $\log_{2x}(4z) = \frac{c + 2}{a + 1}$ ， $\log_{2x^4}(8yz) = \frac{b + c + 3}{4a + 1}$ ，条件变为 $\begin{aligned} \frac{2(b + 1)}{a} &= \frac{2(c + 2)}{a + 1} \\ &= \frac{b + c + 3}{4a + 1} \ne 0 \end{aligned}\text{。}$

由前两个式子， $\frac{b + 1}{a} = \frac{c + 2}{a + 1}$ ，所以这个公共比值也等于分子之和与分母之和的比值 $\frac{b + c + 3}{2a + 1}$ 。与第三个表达式比较，得到 $\frac{2(b + c + 3)}{2a + 1} = \frac{b + c + 3}{4a + 1}$ 。公共值非零，所以 $b + c + 3 \ne 0$ ，从而 $2(4a + 1) = 2a + 1$ ，得到 $a = -\frac{1}{6}$ 。又由 $\frac{b + 1}{-\frac{1}{6}} = \frac{c + 2}{\frac{5}{6}}$ ，得 $c + 2 = -5(b + 1)$ ，即 $5b + c = -7$ 。

因此 $xy^5z = 2^{a + 5b + c} = 2^{-\frac{1}{6} - 7} = \frac{1}{2^{\frac{43}{6}}}$ ，所以 $p + q = 43 + 6 = 49$ 。

Write $x = 2^a,$ $y = 2^b,$ $z = 2^c.$ Then $\log_x(2y) = \frac{b + 1}{a},$ $\log_{2x}(4z) = \frac{c + 2}{a + 1},$ and $\log_{2x^4}(8yz) = \frac{b + c + 3}{4a + 1},$ so the condition is $\begin{aligned} \frac{2(b + 1)}{a} &= \frac{2(c + 2)}{a + 1} \\ &= \frac{b + c + 3}{4a + 1} \ne 0. \end{aligned}$

From the first two, $\frac{b + 1}{a} = \frac{c + 2}{a + 1},$ and equal ratios also equal their mediant $\frac{b + c + 3}{2a + 1}.$ Comparing with the third expression gives $\frac{2(b + c + 3)}{2a + 1} = \frac{b + c + 3}{4a + 1}.$ The common value is nonzero, so $b + c + 3 \ne 0,$ and thus $2(4a + 1) = 2a + 1,$ giving $a = -\frac{1}{6}.$ Then $\frac{b + 1}{-\frac{1}{6}} = \frac{c + 2}{\frac{5}{6}}$ yields $c + 2 = -5(b + 1),$ that is, $5b + c = -7.$

Therefore $xy^5z = 2^{a + 5b + c} = 2^{-\frac{1}{6} - 7} = \frac{1}{2^{\frac{43}{6}}},$ so $p + q = 43 + 6 = 49.$

2012 AIME I 第 9 题

答案：49

难度评级：2650

小提示：

大提示：

解答：

其他年份的第 9 题