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

Revision as of 17:58, 18 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 1

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)

Solution 2

We have $\begin{align*} \log_{20x} (22x) & = \frac{\log_k 22x}{\log_k 20x} \\ & = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . \end{align*}$

We have $\begin{align*} \log_{2x} (202x) & = \frac{\log_k 202x}{\log_k 2x} \\ & = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \end{align*}$

Because $\log_{20x} (22x)=\log_{2x} (202x)$ , we get $\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} .$

We denote this common value as $\lambda$ .

By solving the equality $\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \lambda$ , we get $\log_k x = \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}$ .

By solving the equality $\frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} = \lambda$ , we get $\log_k x = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1}$ .

By equating these two equations, we get $\frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1} = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1} .$

Therefore, $\begin{align*} \log_{20x} (22x) & = \lambda \\ & = \frac{\log_k 22 - \log_k 202}{\log_k 20 - \log_k 2} \\ & = \frac{\log_k \frac{11}{101}}{\log_k 10} \\ & = \log_{10} \frac{11}{101} . \end{align*}$

Therefore, the answer is $11 + 101 = \boxed{\textbf{(112) }}$ .

~Steven Chen (www.professorchenedu.com)

Video Solution

https://www.youtube.com/watch?v=4qJyvyZN630

@@ Line 75: / Line 75: @@
 ~Steven Chen (www.professorchenedu.com)
+==Video Solution==
+https://www.youtube.com/watch?v=4qJyvyZN630
 ==See Also==
 {{AIME box|year=2022|n=II|num-b=3|num-a=5}}
 {{MAA Notice}}

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