2022 AIME II Problem 4

Change both sides to a common base: $\frac{\ln 22x}{\ln 20x} = \frac{\ln 202x}{\ln 2x}$

If $\frac{p}{q} = \frac{r}{s},$ then both fractions also equal $\frac{r - p}{s - q}.$ Apply this and simplify the resulting quotient.

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: $$\begin{aligned} y &= \frac{\ln 202x - \ln 22x}{\ln 2x - \ln 20x} \\ &= \frac{\ln \frac{101}{11}}{\ln \frac{1}{10}} \\ &= -\log_{10}\frac{101}{11} \\ &= \log_{10}\frac{11}{101}. \end{aligned}$$

To check existence rather than merely use the promised $x,$ note that $y \ne 1.$ Setting $$\ln x=\frac{y\ln 20-\ln 22}{1-y}$$ makes $\ln(22x)=y\ln(20x).$ Also $$\begin{aligned} \ln(202x)-\ln(22x) &= \ln\frac{101}{11} \\ &= y\ln\frac{1}{10}, \end{aligned}$$ so $\ln(202x)=y\ln(2x)$ as well. This positive $x$ is neither excluded value (neither one satisfies the displayed linear equation), so both logarithm bases are valid. Since $\gcd(11,101)=1,$ we get $m+n=11+101=112.$