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 1

Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$ , what we are looking for. Then, by the definition of the logarithm, $\begin{cases} (20x)^{a} &= 22x \\ (2x)^{a} &= 202x. \end{cases}$ Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$ , so by the definition of logs, $a = \log_{10} \frac{11}{101}$ . This is what the problem asked for, so the fraction $\frac{11}{101}$ gives us $m+n = \boxed{112}$ .

~ihatemath123

Solution 2

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: $(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot \left(10^v\right)=22x \textcircled{3}$

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

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

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

Solution 3

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)

Solution 4 (Solution 1 with more reasoning)

Let $a$ be the exponent such that $(20x)^a = 22x$ and $(2x)^a = 202x$ . Dividing, we get $\begin{align*} \dfrac{(20x)^a}{(2x)^a} &= \dfrac{22x}{202x}. \\ \left(\dfrac{20x}{2x}\right)^a &= \dfrac{22x}{202x}. \\ 10^a &= \dfrac{11}{101}. \\ \end{align*}$ Thus, we see that $\log_{10} \left(\dfrac{11}{101}\right) = a = \log_{20x} 22x$ , so the answer is $11 + 101 = \boxed{112}$ .

~A_MatheMagician

Solution 5

By the change of base rule, we have $\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}$ , or $\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k$ . We also know that if $a/b=c/d$ , then this also equals $\frac{a-c}{b-d}$ . We use this identity and find that $k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{22}=\log\frac{11}{101}$ . The requested sum is $11+101=\boxed{112}.$

~MathIsFun286

Video Solution

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

Video Solution by Power of Logic

https://youtu.be/m2Cm9r5_Jvs

~Hayabusa1

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