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

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 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

See Also

2022 AIME II (ProblemsAnswer KeyResources)
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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