AMC12 2023 A

AMC12 2023 A · Q19

AMC12 2023 A · Q19. It mainly tests Vieta / quadratic relationships (basic), Logarithms (rare).

What is the product of all solutions to the equation \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]

方程 \[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\] 的所有解的乘积是多少？

(A) (\log_{2023}7\cdot \log_{2023}289)^2 (\log_{2023}7\cdot \log_{2023}289)^2

(B) \log_{2023}7\cdot \log_{2023}289 \log_{2023}7\cdot \log_{2023}289

(D) \log_{7}2023\cdot \log_{289}2023 \log_{7}2023\cdot \log_{289}2023

(E) (\log_7 2023\cdot\log_{289} 2023)^2 (\log_7 2023\cdot\log_{289} 2023)^2

Answer

Correct choice: (C)

正确答案：（C）

Solution

For $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$, transform it into $\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$. Replace $\ln x$ with $y$. Because we want to find the product of all solutions of $x$, it is equivalent to finding the exponential of the sum of all solutions of $y$. Change the equation to standard quadratic equation form, the term with 1 power of $y$ is canceled. By using Vieta, we see that since there does not exist a $by$ term, $\sum y=0$ and $\prod x=e^0=\boxed{\textbf{(C)} 1}$.

对于 $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$，将其转化为 $\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$。将 $\ln x$ 替换为 $y$。因为我们想要找到所有 $x$ 解的乘积，这等价于找到所有 $y$ 解之和的指数。将方程改为标准二次方程形式，一次项的系数为零。利用 Vieta 定理，我们看到由于不存在 $by$ 项，$\sum y=0$，且 $\prod x=e^0=\boxed{\textbf{(C)} 1}$。

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