AIME 2024b Problem 4 PaperAnswer Let x,yx,yx,y and zzz be positive real numbers that satisfy the following system of equations: log2(xyz)=12\log_2\left({x \over yz}\right) = {1 \over 2}log2(yzx)=21log2(yxz)=13\log_2\left({y \over xz}\right) = {1 \over 3}log2(xzy)=31log2(zxy)=14\log_2\left({z \over xy}\right) = {1 \over 4}log2(xyz)=41 Then the value of ∣log2(x4y3z2)∣\left|\log_2(x^4y^3z^2)\right|log2(x4y3z2) is mn\tfrac{m}{n}nm where mmm and nnn are relatively prime positive integers. Find m+nm+nm+n. Dataset text AIME 1983 to 2026. Quoted with source. Show the answer 33 Solutions No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label. Add a solution PreviousProblem 3 NextProblem 5 Locator aime/2024b/p4 · Tag Z6SENYPNJA · Edit this page on GitHub · Keys: j next, k previous