2024 AIME II Problems/Problem 4

Problem

Let $x,y$ and $z$ be positive real numbers that satisfy the following system of equations: $\log_2\left({x \over yz}\right) = {1 \over 2}$ $\log_2\left({y \over xz}\right) = {1 \over 3}$ $\log_2\left({z \over xy}\right) = {1 \over 4}$ Then the value of $\left|\log_2(x^4y^3z^2)\right|$ is ${m \over n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

Denote $\log_2(x) = a$ , $\log_2(y) = b$ , and $\log_2(z) = c$ .

Then, we have: $a-b-c = \frac{1}{2}$ $-a+b-c = \frac{1}{3}$ $-a-b+c = \frac{1}{4}$

Now, we can solve to get $a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}$ . Plugging these values in, we obtain $\mid 4a + 3b + 2c \mid = \frac{25}{8} \implies \boxed{033}$ . ~akliu

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2024 AIME II Problems/Problem 4

Problem

Solution 1