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Difference between revisions of "2024 AIME I Problems/Problem 2"

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There exist real numbers <math>x</math> and <math>y</math>, both greater than 1, such that <math>\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10</math>. Find <math>xy</math>.  
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==Problem==
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There exist real numbers <math>x</math> and <math>y</math>, both greater than 1, such that <math>\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10</math>. Find <math>xy</math>.
 +
 
 +
==Video Solution & More by MegaMath==
 +
https://www.youtube.com/watch?v=jxY7BBe-4gU
 +
 
 +
==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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\begin{align*}
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x\log_xy&=10 \\
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<cmath>x\log_xy=10</cmath>
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4y\log_yx&=10. \\
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<cmath>4y\log_yx=10.</cmath>
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\end{align*}
 
 
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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~Technodoggo
 
~Technodoggo
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 +
==Solution 2==
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 +
Convert the two equations into exponents:
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 +
<cmath>x^{10}=y^x~(1)</cmath>
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<cmath>y^{10}=x^{4y}~(2).</cmath>
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Take <math>(1)</math> to the power of <math>\frac{1}{x}</math>:
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<cmath>x^{\frac{10}{x}}=y.</cmath>
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Plug this into <math>(2)</math>:
 +
 +
<cmath>x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}</cmath>
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<cmath>{\frac{100}{x}}={4x^{\frac{10}{x}}}</cmath>
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<cmath>{\frac{25}{x}}={x^{\frac{10}{x}}}=y,</cmath>
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 +
So <math>xy=\boxed{025}</math>
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 +
~alexanderruan
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==Solution 3==
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 +
Similar to solution 2, we have:
 +
 +
<math>x^{10}=y^x</math> and <math>y^{10}=x^{4y}</math>
 +
 +
Take the tenth root of the first equation to get
 +
 +
<math>x=y^{\frac{x}{10}}</math>
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 +
Substitute into the second equation to get
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<math>y^{10}=y^{\frac{4xy}{10}}</math>
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This means that <math>10=\frac{4xy}{10}</math>, or <math>100=4xy</math>, meaning that <math>xy=\boxed{025}</math>.
 +
 +
~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>.So <math>xy = \boxed{025}</math>.
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==Solution 5==
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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{025}</math>.
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 +
~idk12345678
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==Solution 6==
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Put <cmath> y=x^a </cmath>.We see: <cmath>ax=10 </cmath> and <cmath>4x^a/a=10 </cmath>
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which gives rise to <cmath>xy = \boxed{025}</cmath>.
 +
 +
~Grammaticus
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 +
==Video Solution==
 +
 +
https://youtu.be/qLUahGcewT4
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Video Solution==
 +
 +
https://youtu.be/6C0yHp5GUBY
 +
 +
~Veer Mahajan
  
 
==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}}
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[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 05:34, 24 November 2024

Problem

There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=jxY7BBe-4gU

Video Solution By MathTutorZhengFrSG

https://youtu.be/HbGlIki_BsY

~MathTutorZhengFrSG

Solution 1

By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$. Let us break this into two separate equations:

\[x\log_xy=10\] \[4y\log_yx=10.\] We multiply the two equations to get: \[4xy\left(\log_xy\log_yx\right)=100.\]

Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$; thus, $\log_xy\cdot\log_yx=1$. Therefore, our equation simplifies to:

\[4xy=100\implies xy=\boxed{025}.\]

~Technodoggo

Solution 2

Convert the two equations into exponents:

\[x^{10}=y^x~(1)\] \[y^{10}=x^{4y}~(2).\]

Take $(1)$ to the power of $\frac{1}{x}$:

\[x^{\frac{10}{x}}=y.\]

Plug this into $(2)$:

\[x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}\] \[{\frac{100}{x}}={4x^{\frac{10}{x}}}\] \[{\frac{25}{x}}={x^{\frac{10}{x}}}=y,\]

So $xy=\boxed{025}$

~alexanderruan

Solution 3

Similar to solution 2, we have:

$x^{10}=y^x$ and $y^{10}=x^{4y}$

Take the tenth root of the first equation to get

$x=y^{\frac{x}{10}}$

Substitute into the second equation to get

$y^{10}=y^{\frac{4xy}{10}}$

This means that $10=\frac{4xy}{10}$, or $100=4xy$, meaning that $xy=\boxed{025}$.

~MC413551

Solution 4

The same with other solutions, we have obtained $x^{10}=y^x$ and $y^{10}=x^{4y}$. Then, $x^{10}y^{10}=y^xx^{4y}$. So, an obvious solution is to have $x^{10}=x^{4y}$ and $y^{10}=y^{x}$. Solving, we get $x=10$ and $y=2.5$.So $xy = \boxed{025}$.

Solution 5

Using the first expression, we see that $x^{10} = y^x$. Now, taking the log of both sides, we get $\log_y(x^{10}) = \log_y(y^x)$. This simplifies to $10 \log_y(x) = x$. This is still equal to the second equation in the problem statement, so $10 \log_y(x) = x = 4y \log_y(x)$. Dividing by $\log_y(x)$ on both sides, we get $x = 4y = 10$. Therefore, $x = 10$ and $y = 2.5$, so $xy = \boxed{025}$.

~idk12345678

Solution 6

Put \[y=x^a\].We see: \[ax=10\] and \[4x^a/a=10\] which gives rise to \[xy = \boxed{025}\].

~Grammaticus

Video Solution

https://youtu.be/qLUahGcewT4

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/6C0yHp5GUBY

~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

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