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 $m+n$.

Let's think step by step and output the final answer within \boxed{}.
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