Difference between revisions of "2023 AMC 12A Problems/Problem 19"
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==Solution 2== | ==Solution 2== | ||
| − | < | + | <cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath> |
Rearranging it give us: | Rearranging it give us: | ||
| − | < | + | <cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x</cmath> |
| − | < | + | <cmath>(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)</cmath> |
let <math>\log_{2023}x</math> be <math>a</math>, we get | let <math>\log_{2023}x</math> be <math>a</math>, we get | ||
| − | < | + | <cmath>(\log_{2023}7+a)(\log_{2023}289+a)=1+a</cmath> |
| − | < | + | <cmath>a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a</cmath> |
| − | < | + | <cmath>a^2+\log_{2023}7 \cdot \log_{2023}289-1=0</cmath> |
by veita's formula, | by veita's formula, | ||
| − | < | + | <cmath>a_1+a_2=0</cmath> |
| − | < | + | <cmath>\log_{2023}{x_1}+\log_{2023}{x_2}=0</cmath> |
| − | < | + | <cmath>\log_{2023}{x_1x_2}=0</cmath> |
| − | < | + | <cmath>x_1x_2=\boxed{\textbf{(C)} 1}</cmath> |
~lptoggled | ~lptoggled | ||
Revision as of 00:10, 10 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
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 veita's formula,
![\[a_1+a_2=0\]](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
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. 
