Difference between revisions of "2023 AMC 12A Problems/Problem 19"
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<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)} | + | <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 \\ \qquad \textbf{(D) } \log_{7}2023\cdot \log_{289}2023\qquad \textbf{(E) } (\log_7 2023\cdot\log_{289} 2023)^2</math> |
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==Solution 1== | ==Solution 1== | ||
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~dwarf_marshmallow | ~dwarf_marshmallow | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ~MLiang2018 | ||
+ | |||
+ | ==Solution 5(Similar to solution 4)== | ||
+ | First, we take the reciprocal of both sides. We get <cmath>\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.</cmath> | ||
+ | Flip the logarithms to get <cmath>\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.</cmath> | ||
+ | |||
+ | Now we can use <math>\log_{a}bc=\log_{a}b+\log_{a}c</math>. We get <cmath>\log_{2023}7+\log_{2023}x+\log_{2023}289+\log_{2023}x=\log_{2023}2023+\log_{2023}x.</cmath> The <math>\log_{2023}7+\log_{2023}289</math> and <math>\log_{2023}2023</math> terms cancel, giving <cmath>2\cdot\log_{2023}x=\log_{2023}x</cmath> so now we are sure that <math>\log_{2023}x=0</math>, so the only solution is <math>x=\boxed{\textbf{(C)} 1}</math>. | ||
+ | |||
+ | ~Yrock | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Latest revision as of 22:22, 2 August 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
~MLiang2018
Solution 5(Similar to solution 4)
First, we take the reciprocal of both sides. We get
Flip the logarithms to get
Now we can use . We get
The
and
terms cancel, giving
so now we are sure that
, so the only solution is
.
~Yrock
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.