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2022 AIME II Problems/Problem 4: Difference between revisions

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There is a positive real number <math>x</math> not equal to either <math>\tfrac{1}{20}</math> or <math>\tfrac{1}{2}</math> such that<cmath>\log_{20x} (22x)=\log_{2x} (202x).</cmath>The value <math>\log_{20x} (22x)</math> can be written as <math>\log_{10} (\tfrac{m}{n})</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
There is a positive real number <math>x</math> not equal to either <math>\tfrac{1}{20}</math> or <math>\tfrac{1}{2}</math> such that<cmath>\log_{20x} (22x)=\log_{2x} (202x).</cmath>The value <math>\log_{20x} (22x)</math> can be written as <math>\log_{10} (\tfrac{m}{n})</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
==Solution 1==
Define <math>a</math> to be <math>\log_{20x} (22x) = \log_{2x} (202x)</math>, what we are looking for. Then, by the definition of logs,
<cmath>\begin{cases}
(20x)^{a} &= 22x \\
(2x)^{a} &= 202x.
\end{cases}</cmath>
Dividing the first equation by the second equation gives us <math>10^a = \frac{11}{101}</math>, so by the definition of logs, <math>a = \log_{10} \frac{11}{101}</math>. This is what the problem asked for, so the fraction <math>\frac{11}{101}</math> gives us <math>m+n = \boxed{112}</math>.


==Solution 1==
~ihatemath123
==Solution 2==
We could assume a variable <math>v</math> which equals to both <math>\log_{20x} (22x)</math> and <math>\log_{2x} (202x)</math>.
We could assume a variable <math>v</math> which equals to both <math>\log_{20x} (22x)</math> and <math>\log_{2x} (202x)</math>.


Line 19: Line 27:
~DSAERF-CALMIT (https://binaryphi.site)
~DSAERF-CALMIT (https://binaryphi.site)


==Solution 2==
==Solution 3==


We have
We have

Revision as of 20:35, 19 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

Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$, what we are looking for. Then, by the definition of logs, \[\begin{cases} (20x)^{a} &= 22x \\ (2x)^{a} &= 202x. \end{cases}\] 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 3

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 $\frac{a-b}{c-d}$. We use this identity and find that $k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{20}=\log\frac{11}{101}$. The requested sum is $11+101=\boxed{112}.$

~MathIsFun286

Video Solution

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

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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