There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that\[\log_{20x} (22x)=\log_{2x} (202x).\]The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$, what we are looking for. Then, by the definition of the logarithm,
\[\begin{cases}  (20x)^{a} &= 22x \\ (2x)^{a} &= 202x.  \end{cases}\]
Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$, so by the definition of logs, $a = \log_{10} \frac{11}{101}$. This is what the problem asked for, so the fraction $\frac{11}{101}$ gives us $m+n = \boxed{112}$.
~ihatemath123