HMMT 二月 2020 · ALGNT 赛 · 第 3 题
HMMT February 2020 — ALGNT Round — Problem 3
题目详情
- Let a = 256. Find the unique real number x > a such that log log log x = log 2 log 2 log 2 x. a a a a a a
解析
- Let a = 256. Find the unique real number x > a such that log log log x = log log log x. 2 2 2 a a a a a a Proposed by: James Lin 32 Answer: 2 1 Solution: Let y = log x so log log y = log 2 log 2 y . Setting z = log y , we find log z = a a a a a a a 2 1 1 2 1 1 1 32 log ( z − ), or z − z + = 0. Thus, we have z = , so we can backsolve to get y = 4 and x = 2 . 2 a 2 16 2 16 4