Question

设 $a$ 、 $b$ 、 $c$ 为正实数，满足 $a^{\log_3 7} = 27$ ， $b^{\log_7 11} = 49$ ，且 $c^{\log_{11} 25} = \sqrt{11}$ 。求 $a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}\text{。}$

Suppose that $a,$ $b,$ and $c$ are positive real numbers such that $a^{\log_3 7} = 27,$ $b^{\log_7 11} = 49,$ and $c^{\log_{11} 25} = \sqrt{11}.$ Find $a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.$

答案

答案：469

知识点：对数指数

难度评级：2150

小提示：

将 $a^{(\log_3 7)^2}$ 写成 $\left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7}$

Write $a^{(\log_3 7)^2}$ as $\left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7}$

大提示：

$27^{\log_3 7} = \left(3^{\log_3 7}\right)^3 = 7^3$ 。另外两项用同样方法处理。

$27^{\log_3 7} = \left(3^{\log_3 7}\right)^3 = 7^3.$ Handle the other two terms the same way.

解答：

由指数的乘方法则， $\begin{aligned} a^{(\log_3 7)^2} &= \left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7} \\ &= \left(3^{\log_3 7}\right)^3 = 7^3 = 343 \end{aligned}\text{。}$

同理， $\begin{aligned} b^{(\log_7 11)^2} &= 49^{\log_7 11} = \left(7^{\log_7 11}\right)^2 \\ &= 11^2 = 121 \end{aligned}\text{，}$ 并且 $\begin{aligned} c^{(\log_{11} 25)^2} &= \left(\sqrt{11}\right)^{\log_{11} 25} \\ &= \left(11^{\log_{11} 25}\right)^{\frac{1}{2}} \\ &= 25^{\frac{1}{2}} = 5 \end{aligned}\text{。}$

所以和为 $343 + 121 + 5 = 469$ 。

By the power rule for exponents, $\begin{aligned} a^{(\log_3 7)^2} &= \left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7} \\ &= \left(3^{\log_3 7}\right)^3 = 7^3 = 343. \end{aligned}$

In the same way, $\begin{aligned} b^{(\log_7 11)^2} &= 49^{\log_7 11} = \left(7^{\log_7 11}\right)^2 \\ &= 11^2 = 121, \end{aligned}$ and $\begin{aligned} c^{(\log_{11} 25)^2} &= \left(\sqrt{11}\right)^{\log_{11} 25} \\ &= \left(11^{\log_{11} 25}\right)^{\frac{1}{2}} \\ &= 25^{\frac{1}{2}} = 5. \end{aligned}$

The sum is $343 + 121 + 5 = 469.$

2009 AIME II 第 2 题

答案：469

难度评级：2150

小提示：

大提示：

解答：

其他年份的第 2 题