Question

There is a positive real number $x$ not equal to either $\frac{1}{20}$ or $\frac{1}{2}$ such that $\log_{20x}(22x) = \log_{2x}(202x).$ The value $\log_{20x}(22x)$ can be written as $\log_{10}\left(\frac{m}{n}\right),$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

Let $y$ be the common value. In natural logarithms, $y = \frac{\ln 22x}{\ln 20x} = \frac{\ln 202x}{\ln 2x}.$ When two fractions are equal, each also equals the quotient of the differences of numerators and denominators: $y = \frac{\ln 202x - \ln 22x}{\ln 2x - \ln 20x} = \frac{\ln \frac{101}{11}}{\ln \frac{1}{10}} = -\log_{10}\frac{101}{11} = \log_{10}\frac{11}{101}.$

(Such an $x$ exists: the equation rearranges to a solvable condition, and the excluded values $\frac{1}{20}, \frac{1}{2}$ only rule out degenerate bases.) Since $\gcd(11, 101) = 1,$ we get $m + n = 11 + 101 = 112.$

2022 AIME II Problem 4

Solution:

Problem 4 in Other Years