Difference between revisions of "2023 AMC 12A Problems/Problem 19"

Revision as of 19:59, 30 September 2024

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 understood 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

This solution works for this problem by chance but do note that the simplification step to get $2023x^2=2023x$ is not how log properties work and that the actual solutions for x are $x = e^{ \pm \sqrt{ln^2(7)+ln(17)ln(2092529)}}$ (as shown in solution 3) which multiply to 1

~silk-hyacinth

Solution 5(Similar to solution 4, Fakesolve)

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 1 by OmegaLearn

https://youtu.be/OcNU62SMh4o

Video Solution

https://youtu.be/-CZkFE-wriQ

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

@@ Line 62: / Line 62: @@
 ~silk-hyacinth
-==Solution 5(Similar to solution 4)==
+==Solution 5(Similar to solution 4, Fakesolve)==
 First, we take the reciprocal of both sides. We get <cmath>\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.</cmath>
 Flip the logarithms to get <cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.</cmath>

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search