Difference between revisions of "2023 AMC 12A Problems/Problem 19"
| Line 3: | Line 3: | ||
<cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath> | <cmath>\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023</cmath> | ||
| − | <math>\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1 | + | <math>\textbf{(A)} ~(\log_{2023}7\cdot \log_{2023}289)^2\qquad\textbf{(B)} ~\log_{2023}7\cdot \log_{2023}289\qquad\textbf{(C)} ~1</math> |
| + | <math>\textbf{(D)} ~\log_{7}2023\cdot \log_{289}2023\qquad\textbf{(E)} ~(\log_7 2023\cdot\log_{289} 2023)^2</math> | ||
Revision as of 19:32, 9 November 2023
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 Veita, we see that since there does not exist a 
term, 
and 
.
~plasta
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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