Difference between revisions of "2022 AIME II Problems/Problem 4"

Revision as of 22:59, 17 February 2022

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

@@ Line 13: / Line 13: @@
 Substitute <math>\textcircled{2}</math> to <math>\textcircled{3}</math>: <math>202x \cdot (10^v)=22x</math>
-Thus, <math>v=\log_{10} (\tfrac{22x}{202x})= \log_{10} (\tfrac{11}{101})</math>, where <math>m=11</math> and <math>n=101</math>.
+Thus, <math>v=\log_{10} (\frac{22x}{202x})= \log_{10} (\frac{11}{101})</math>, where <math>m=11</math> and <math>n=101</math>.
 Therefore, <math>m+n = \boxed{112}</math>.

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Difference between revisions of "2022 AIME II Problems/Problem 4"

Revision as of 22:59, 17 February 2022

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