2022 AIME II Problems/Problem 4

Problem

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 $m+n$ .

Solution

We could assume a variable $v$ which equals to both $\log_{20x} (22x)$ and $\log_{2x} (202x)$ .

So that $(20x)^v=22x \textcircled{1}$ and $(2x)^v=202x \textcircled{2}$

Express $\textcircled{1}$ as: $(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x \textcircled{3}$

Substitute $\textcircled{2}$ to $\textcircled{3}$ : $202x \cdot (10^v)=22x$

Thus, $v=\log_{10} (\frac{22x}{202x})= \log_{10} (\frac{11}{101})$ , where $m=11$ and $n=101$ .

Therefore, $m+n = \boxed{112}$ .

~DSAERF-CALMIT (https://binaryphi.site)

2022 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2022 AIME II Problems/Problem 4

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