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
[hide]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:
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
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.