Difference between revisions of "2023 AMC 12A Problems/Problem 19"
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+ | ==Solution 4== | ||
+ | We take the reciprocal of both sides: | ||
+ | <cmath>\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.</cmath> Using logarithm properties, we have <cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.</cmath> Simplify to obtain <cmath>2023x^2=2023x,</cmath> from which we have <math>x=\boxed{\textbf{(C)} 1}</math> | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 00:43, 29 February 2024
Contents
Problem
What is the product of all solutions to the equation
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:
let be , we get
by Vieta's Formulas,
~lptoggled
Solution 3
Similar to solution 1, change the bases first Cancel and cross multiply to get Simplify to get 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: Using logarithm properties, we have Simplify to obtain from which we have
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.