Difference between revisions of "2024 AIME I Problems/Problem 2"
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==Video Solution & More by MegaMath== | ==Video Solution & More by MegaMath== | ||
https://www.youtube.com/watch?v=jxY7BBe-4gU | https://www.youtube.com/watch?v=jxY7BBe-4gU | ||
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+ | ==Video Solution By MathTutorZhengFrSG== | ||
+ | |||
+ | https://youtu.be/HbGlIki_BsY | ||
+ | |||
+ | ~MathTutorZhengFrSG | ||
==Solution 1== | ==Solution 1== | ||
By properties of logarithms, we can simplify the given equation to <math>x\log_xy=4y\log_yx=10</math>. Let us break this into two separate equations: | By properties of logarithms, we can simplify the given equation to <math>x\log_xy=4y\log_yx=10</math>. Let us break this into two separate equations: | ||
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− | <cmath>x\log_xy | + | <cmath>x\log_xy=10</cmath> |
− | <cmath>4y\log_yx | + | <cmath>4y\log_yx=10.</cmath> |
We multiply the two equations to get: | We multiply the two equations to get: | ||
<cmath>4xy\left(\log_xy\log_yx\right)=100.</cmath> | <cmath>4xy\left(\log_xy\log_yx\right)=100.</cmath> | ||
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This means that <math>10=\frac{4xy}{10}</math>, or <math>100=4xy</math>, meaning that <math>xy=\boxed{25}</math>. | This means that <math>10=\frac{4xy}{10}</math>, or <math>100=4xy</math>, meaning that <math>xy=\boxed{25}</math>. | ||
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~MC413551 | ~MC413551 | ||
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+ | ==Solution 4== | ||
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+ | The same with other solutions, we have obtained <math>x^{10}=y^x</math> and <math>y^{10}=x^{4y}</math>. Then, <math>x^{10}y^{10}=y^xx^{4y}</math>. So, an obvious solution is to have <math>x^{10}=x^{4y}</math> and <math>y^{10}=y^{x}</math>. Solving, we get <math>x=10</math> and <math>y=2.5</math>. | ||
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+ | ==Solution 5== | ||
+ | Using the first expression, we see that <math>x^{10} = y^x</math>. Now, taking the log of both sides, we get <math>\log_y(x^{10}) = \log_y(y^x)</math>. This simplifies to <math>10 \log_y(x) = x</math>. This is still equal to the second equation in the problem statement, so <math>10 \log_y(x) = x = 4y \log_y(x)</math>. Dividing by <math>\log_y(x)</math> on both sides, we get <math>x = 4y = 10</math>. Therefore, <math>x = 10</math> and <math>y = 2.5</math>, so <math>xy = \boxed{25}</math>. | ||
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+ | ~idk12345678 | ||
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+ | ==Solution 6== | ||
+ | Put <cmath> y=x^a </cmath>.We see: <cmath>ax=10 </cmath> and <cmath>4x^a/a=10 </cmath> | ||
+ | which gives rise to <cmath>xy=25 </cmath> which is the required answer. | ||
+ | |||
+ | ~Grammaticus | ||
==Video Solution== | ==Video Solution== | ||
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==See also== | ==See also== | ||
{{AIME box|year=2024|n=I|num-b=1|num-a=3}} | {{AIME box|year=2024|n=I|num-b=1|num-a=3}} | ||
− | + | [[Category:Introductory Algebra Problems]] | |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 06:35, 16 November 2024
Contents
[hide]Problem
There exist real numbers and , both greater than 1, such that . Find .
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=jxY7BBe-4gU
Video Solution By MathTutorZhengFrSG
~MathTutorZhengFrSG
Solution 1
By properties of logarithms, we can simplify the given equation to . Let us break this into two separate equations:
We multiply the two equations to get:
Also by properties of logarithms, we know that ; thus, . Therefore, our equation simplifies to:
~Technodoggo
Solution 2
Convert the two equations into exponents:
Take to the power of :
Plug this into :
So
~alexanderruan
Solution 3
Similar to solution 2, we have:
and
Take the tenth root of the first equation to get
Substitute into the second equation to get
This means that , or , meaning that .
~MC413551
Solution 4
The same with other solutions, we have obtained and . Then, . So, an obvious solution is to have and . Solving, we get and .
Solution 5
Using the first expression, we see that . Now, taking the log of both sides, we get . This simplifies to . This is still equal to the second equation in the problem statement, so . Dividing by on both sides, we get . Therefore, and , so .
~idk12345678
Solution 6
Put .We see: and which gives rise to which is the required answer.
~Grammaticus
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Veer Mahajan
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.