Question 195
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 $\log_{10}(\frac{m}{n})$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
$$
\log_{20x}(22x) = \log_{2x}(202x).
$$
The value $\log_{20x}(22x)$ can be written as $\log_{10}(\frac{m}{n})$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.