PUMaC 2016 · 代数（B 组） · 第 5 题

PUMaC 2016 — Algebra (Division B) — Problem 5

For positive real numbers x and y , let f ( x, y ) = x . The sum of the solutions to the equation 13 4096 f ( f ( x, x ) , x ) = x m can be written in simplest form as . Compute m + n . n

解析

We have 13 4096 f ( f ( x, x ) , x ) = x ( ) log x 2 log x 13 2 4096 x = x 2 (log x ) 13 2 4096 x = x 2 (log x ) − 13 2 4096 x = 1 2 (log x )((log x ) − 13)+12 0 2 2 2 = 2 2 (log x )((log x ) − 13) + 12 = 0 2 2 3 (log x ) − 13 log x + 12 = 0 2 2 (log x − 1)(log x − 3)(log x + 4) = 0 , 2 2 2 1 161 so x can be any of 2, 8, and . Thus, the sum of all possible values of x is , and so our 16 16 answer is 161 + 16 = 177 . Problem written by Eric Neyman.