Question

Let $x,$ $y,$ and $z$ be positive real numbers that satisfy the following system of equations: $\log_2\left(\frac{x}{yz}\right) = \frac{1}{2}$ $\log_2\left(\frac{y}{xz}\right) = \frac{1}{3}$ $\log_2\left(\frac{z}{xy}\right) = \frac{1}{4}$

Then the value of $\left|\log_2(x^4 y^3 z^2)\right|$ is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

Let $a = \log_2 x,$ $b = \log_2 y,$ $c = \log_2 z.$ The equations become $a - b - c = \frac{1}{2},$ $b - a - c = \frac{1}{3},$ $c - a - b = \frac{1}{4}.$ Adding all three gives $-(a + b + c) = \frac{13}{12}.$ Since $a - b - c = 2a - (a + b + c),$ we get $2a = \frac{1}{2} - \frac{13}{12} = -\frac{7}{12},$ so $a = -\frac{7}{24},$ and similarly $b = -\frac{3}{8}$ and $c = -\frac{5}{12}.$

Therefore $4a + 3b + 2c = -\frac{28}{24} - \frac{27}{24} - \frac{20}{24} = -\frac{75}{24} = -\frac{25}{8},$ so $\left|\log_2(x^4 y^3 z^2)\right| = \frac{25}{8}$ and $m + n = 25 + 8 = 33.$

2024 AIME II Problem 4

Solution:

Problem 4 in Other Years