2024 AIME I Problem 2

Question

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.

Answer

Bring the exponents down:

x \log_x y = 10

and

4y \log_y x = 10

Multiply the two equations: since

\log_x y \cdot \log_y x = 1,

the logarithms disappear

Pulling the exponents out of the logarithms, the conditions become

x \log_x y = 10

and

4y \log_y x = 10.

Multiplying these equations and using

\log_x y \cdot \log_y x = 1

gives

4xy = 100,

so

xy = 25.

Such

x

and

y

do exist. Set

y = \frac{25}{x}.

The continuous function

x\log_x(\frac{25}{x})

tends to infinity as

x \to 1^+,

while at

x = 2

it is less than

10.

Hence it equals

10

for some

1 \lt x \lt 2,

where

y \gt 1.

Problem 2 in other years