Difference between revisions of "2023 AMC 12A Problems/Problem 19"
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Maa is a troll, trying to exploit your cleverness to waste your time. Just select the simplest answer. C. It’s just 1 rather than some complicated expression full of logs and huge numbers. | Maa is a troll, trying to exploit your cleverness to waste your time. Just select the simplest answer. C. It’s just 1 rather than some complicated expression full of logs and huge numbers. | ||
| + | |||
| + | There are plenty of AMC problems that have simple answers which are not correct. Similar to what numerophile said on problem 17. ~andliu766 | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== | ||
Revision as of 17:02, 11 November 2023
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 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
Maa is a troll, trying to exploit your cleverness to waste your time. Just select the simplest answer. C. It’s just 1 rather than some complicated expression full of logs and huge numbers.
There are plenty of AMC problems that have simple answers which are not correct. Similar to what numerophile said on problem 17. ~andliu766
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
