Difference between revisions of "2024 AIME I Problems/Problem 2"
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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=10</cmath> | <cmath>x\log_xy=10</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>. | ||
~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, <math>x^{10}=x^{4y}</math>, <math>y^{10}=y^{x}</math>. <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. | ||
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+ | ~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}} |
Latest revision as of 12:50, 5 June 2024
Contents
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
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,
,
.
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.