# 2023 AMC 12A Problems/Problem 19

## Problem

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

$\textbf{(A) } (\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B) } \log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C) } 1 \\ \qquad \textbf{(D) } \log_{7}2023\cdot \log_{289}2023\qquad \textbf{(E) } (\log_7 2023\cdot\log_{289} 2023)^2$

## Solution 1

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 $\prod x=e^0=\boxed{\textbf{(C)} 1}$.

~plasta

## Solution 2 (Same idea as Solution 1 with easily understand steps)

$$\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$$

Rearranging it give us:

$$\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x$$

$$(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)$$

let $\log_{2023}x$ be $a$, we get

$$(\log_{2023}7+a)(\log_{2023}289+a)=1+a$$

$$a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a$$

$$a^2+\log_{2023}7 \cdot \log_{2023}289-1=0$$

by Vieta's Formulas,

$$a_1+a_2=0$$

$$\log_{2023}{x_1}+\log_{2023}{x_2}=0$$

$$\log_{2023}{x_1x_2}=0$$

$$x_1x_2=\boxed{\textbf{(C)} 1}$$

~lptoggled

## Solution 3

Similar to solution 1, change the bases first $$\frac{\ln 289+\ln 7}{\ln7 + \ln{x}} \cdot \frac{\ln 289+\ln 7}{2\ln17 + \ln{x}} = \frac{\ln 289+\ln 7}{\ln7 + 2\ln17 + \ln{x}}$$ Cancel and cross multiply to get $$(\ln7 + 2\ln17)(\ln7 + 2\ln17 + \ln{x}) = (\ln7 + \ln{x})(2\ln17 + \ln{x})$$ Simplify to get $$(\ln{x})^2 = 4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2$$ $$\ln{x} = \pm \sqrt{4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2}$$ The sum of all possible $\ln{x}$ is 0, thus the product of all solutions of $x$ is $\boxed{\textbf{(C)} 1}$

~dwarf_marshmallow

## Solution 4

We take the reciprocal of both sides: $$\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.$$ Using logarithm properties, we have $$\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.$$ Simplify to obtain $$2023x^2=2023x,$$ from which we have $x=\boxed{\textbf{(C)} 1}$

~MLiang2018

## Solution 5(Similar to solution 4)

First, we take the reciprocal of both sides. We get $$\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.$$ Flip the logarithms to get $$\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.$$

Now we can use $\log_{a}bc=\log_{a}b+\log_{a}c$. We get $$\log_{2023}7+\log_{2023}x+\log_{2023}289+\log_{2023}x=\log_{2023}2023+\log_{2023}x.$$ The $\log_{2023}7+\log_{2023}289$ and $\log_{2023}2023$ terms cancel, giving $$2\cdot\log_{2023}x=\log_{2023}x$$ so now we are sure that $\log_{2023}x=0$, so the only solution is $x=\boxed{\textbf{(C)} 1}$.

~Yrock

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)