2023 AMC 12A Problems/Problem 19
Contents
Problem
What is the product of all solutions to the equation
![\[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]](http://latex.artofproblemsolving.com/4/4/b/44b223caebbcdf31d72ebe8a7c0a02579fc1b63a.png)
Solution 1
For 
, transform it into 
. Replace 
with 
. Because we want to find the product of all solutions of 
, it is equivalent to finding the exponential of the sum of all solutions of . Change the equation to standard quadratic equation form, the term with 1 power of is canceled. By using Vieta, we see that since there does not exist a 
term, 
and 
.
~plasta
Solution 2 (Same idea as Solution 1 with easily understand steps)
Rearranging it give us:
![\[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x\]](http://latex.artofproblemsolving.com/4/e/c/4ecd9259355ae2da41b0be8d462acfc512f7c93c.png)
![\[(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)\]](http://latex.artofproblemsolving.com/9/1/1/911eb664521597afdf7542f30f914abff35a09d2.png)
let 
be 
, we get
![\[(\log_{2023}7+a)(\log_{2023}289+a)=1+a\]](http://latex.artofproblemsolving.com/0/9/9/0997e4c0a5e53d123efec9f18a25b4a18fce7cf0.png)
![\[a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a\]](http://latex.artofproblemsolving.com/9/d/d/9dd33a7ee2dc9e0ddc02efd459c3f2414f01e9df.png)
![\[a^2+\log_{2023}7 \cdot \log_{2023}289-1=0\]](http://latex.artofproblemsolving.com/8/d/e/8dea6c0706093a18b3df6296da809de441a7e0d0.png)
by Vieta's Formulas,
![\[a_1+a_2=0\]](http://latex.artofproblemsolving.com/c/9/3/c93650e559bb0c8a06f07a30e8c91989a28404b7.png)
![\[\log_{2023}{x_1}+\log_{2023}{x_2}=0\]](http://latex.artofproblemsolving.com/d/b/a/dbacf135836e96d33b8e1a1d9ba1b4cc10e16307.png)
![\[\log_{2023}{x_1x_2}=0\]](http://latex.artofproblemsolving.com/9/6/c/96cf347db849d0ce484fb0854e7dbbc0d14e3863.png)
![\[x_1x_2=\boxed{\textbf{(C)} 1}\]](http://latex.artofproblemsolving.com/5/8/6/586aac4e4fc241b92d21740ba820dcd2edcc2a76.png)
~lptoggled
Solution 3
Similar to solution 1, change the bases first
![\[\frac{\ln 289+\ln 7}{\ln7 + \ln{x}} \cdot \frac{\ln 289+\ln 7}{2\ln17 + \ln{x}} = \frac{\ln 289+\ln 7}{\ln7 + 2\ln17 + \ln{x}}\]](http://latex.artofproblemsolving.com/a/6/9/a699dc7f71ee0c4a85d41c15ef1409332f63343c.png)
Cancel and cross multiply to get
![\[(\ln7 + 2\ln17)(\ln7 + 2\ln17 + \ln{x}) = (\ln7 + \ln{x})(2\ln17 + \ln{x})\]](http://latex.artofproblemsolving.com/a/b/2/ab2a7aea06b85b6f66b161f99cdde6d033bddd58.png)
Simplify to get
![\[(\ln{x})^2 = 4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2\]](http://latex.artofproblemsolving.com/7/6/b/76bb681d7f31acc7899183478ec52333eb93dab4.png)
![\[\ln{x} = \pm \sqrt{4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2}\]](http://latex.artofproblemsolving.com/a/0/7/a0754dcd386b7c4f13ba635793822ec6aa630252.png)
The sum of all possible 
is 0, thus the product of all solutions of is 
~dwarf_marshmallow
Solution 4
We take the reciprocal of both sides:
![\[\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.\]](http://latex.artofproblemsolving.com/2/6/c/26c3b1b293e4739524bbe1f2d28c3572ed75e91b.png)
Using logarithm properties, we have ![\[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.\]](http://latex.artofproblemsolving.com/1/5/a/15a538b41f43a2784127faac6b91a491b027657d.png)
Simplify to obtain ![\[2023x^2=2023x,\]](http://latex.artofproblemsolving.com/d/e/1/de11d13b3e2c23de7d7ab8f99c87b26612a7fa8f.png)
from which we have 
~MLiang2018
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
| 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 | |
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