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February 17, 2022
AIME II

Problem Statement

There is a positive real number xx not equal to either 120\tfrac{1}{20} or 12\tfrac{1}{2} such that log20x(22x)=log2x(202x).\log_{20x} (22x)=\log_{2x} (202x). The value log20x(22x)\log_{20x} (22x) can be written as log10(mn)\log_{10} (\tfrac{m}{n}), where mm and nn are relatively prime positive integers. Find m+nm+n.