2024 AIME I Problem 2

Attempt Problem 2 of the 2024 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AIME I solutions, or check the answer key.

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2.

There exist real numbers xx and y,y, both greater than 1,1, such that logx(yx)=logy(x4y)=10.\log_x\left(y^x\right) = \log_y\left(x^{4y}\right) = 10. Find xy.xy.

Answer: 25
Concepts:logarithmalgebraic manipulation
Difficulty rating: 2070
Solution:

Pulling the exponents out of the logarithms, the conditions become xlogxy=10 x \log_x y = 10 and 4ylogyx=10. 4y \log_y x = 10. Multiplying these equations and using logxylogyx=1\log_x y \cdot \log_y x = 1 gives 4xy=100,4xy = 100, so xy=25.xy = 25.

Such xx and yy do exist: the system solves to logxy=10x\log_x y = \frac{10}{x} with y=25x,y = \frac{25}{x}, which has a solution with x,y>1,x, y \gt 1, so the answer is 25.25.

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