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Difference between revisions of "2022 AIME II Problems/Problem 4"
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==Solution==
 
==Solution==
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We could assume a variable <math>v</math> which equals to both <math>\log_{20x} (22x)</math> and <math>\log_{2x} (202x)</math>.
 +
 +
So that <math>(20x)^v=22x </math> (Eq1)
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and <math>(2x)^v=202x </math> (Eq2)
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 +
Express Eq1 as: <math>(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x </math> (Eq3)
 +
 +
Substitute Eq2 to Eq3: <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>.
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Therefore, <math>m+n = \boxed{112}</math>.
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~DSAERF-CALMIT
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2022|n=II|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:33, 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 $(20x)^v=22x$ (Eq1) and $(2x)^v=202x$ (Eq2)

Express Eq1 as: $(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x$ (Eq3)

Substitute Eq2 to Eq3: $202x \cdot (10^v)=22x$

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

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

~DSAERF-CALMIT

See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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