Difference between revisions of "2024 AIME I Problems/Problem 2"
Veerkmahajan (talk | contribs) m (→Video Solution) |
|||
| Line 31: | Line 31: | ||
Plug this into <math>(2)</math>: | Plug this into <math>(2)</math>: | ||
| − | <cmath>x^{(\frac{10}{x})(10)}=x^{ | + | <cmath>x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}</cmath> |
| − | <cmath>{\frac{100}{x}}={ | + | <cmath>{\frac{100}{x}}={4x^{\frac{10}{x}}}</cmath> |
<cmath>{\frac{25}{x}}={x^{\frac{10}{x}}}=y,</cmath> | <cmath>{\frac{25}{x}}={x^{\frac{10}{x}}}=y,</cmath> | ||
Revision as of 22:03, 4 February 2024
Contents
Problem
There exist real numbers 
and 
, both greater than 1, such that 
. Find 
.
Solution 1
By properties of logarithms, we can simplify the given equation to 
. Let us break this into two separate equations:
\begin{align*}
x\log_xy&=10 \\
4y\log_yx&=10. \\
\end{align*}
We multiply the two equations to get:
![\[4xy\left(\log_xy\log_yx\right)=100.\]](http://latex.artofproblemsolving.com/4/3/3/433f42f4fda44557d99bd60f5db220c4e0682a93.png)
Also by properties of logarithms, we know that 
; thus, 
. Therefore, our equation simplifies to:
![\[4xy=100\implies xy=\boxed{025}.\]](http://latex.artofproblemsolving.com/9/6/8/968ba9e7f231134a15dd1b22aca223a5d5d3947c.png)
~Technodoggo
Solution 2
Convert the two equations into exponents:
![\[x^{10}=y^x~(1)\]](http://latex.artofproblemsolving.com/b/f/2/bf2af4ec313d2cf6f52c0893f418be33bd47d8bd.png)
![\[y^{10}=x^{4y}~(2).\]](http://latex.artofproblemsolving.com/2/9/1/29142adda91dae7f81e9e988e12d5b0380e60fb4.png)
Take 
to the power of 
:
![\[x^{\frac{10}{x}}=y.\]](http://latex.artofproblemsolving.com/3/0/9/3095267bc5f7c2f59735b250a03277b63b994697.png)
Plug this into 
:
![\[x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}\]](http://latex.artofproblemsolving.com/1/d/6/1d6fb0629c4d87e36efa435c6628275423711151.png)
![\[{\frac{100}{x}}={4x^{\frac{10}{x}}}\]](http://latex.artofproblemsolving.com/a/2/5/a250ac7a3cdf8769d70835b33c4c0cbda19e19dc.png)
![\[{\frac{25}{x}}={x^{\frac{10}{x}}}=y,\]](http://latex.artofproblemsolving.com/5/6/3/563826c31e726fe406ecfa632e6635f3ee945c42.png)
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
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. 
