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

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and <math>(2x)^v=202x \textcircled{2}</math>
 
and <math>(2x)^v=202x \textcircled{2}</math>
  
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Express <math>\textcircled{1}</math> as: <math>(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x \textcircled{3}</math>
+
Express <math>\textcircled{1}</math> as: <math>(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot \left(10^v\right)=22x \textcircled{3}</math>
  
−
Substitute <math>\textcircled{2}</math> to <math>\textcircled{3}</math>: <math>202x \cdot (10^v)=22x</math>
+
Substitute <math>\textcircled{{2}}</math> to <math>\textcircled{3}</math>: <math>202x \cdot (10^v)=22x</math>
  
−
Thus, <math>v=\log_{10} (\frac{22x}{202x})= \log_{10} (\frac{11}{101})</math>, where <math>m=11</math> and <math>n=101</math>.
+
Thus, <math>v=\log_{10} \left(\frac{22x}{202x}\right)= \log_{10} \left(\frac{11}{101}\right)</math>, where <math>m=11</math> and <math>n=101</math>.
  
 
Therefore, <math>m+n = \boxed{112}</math>.
 
Therefore, <math>m+n = \boxed{112}</math>.
−
 
−
~DSAERF-CALMIT (https://binaryphi.site)
 
  
 
==Solution 3==
 
==Solution 3==
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~MathIsFun286
 
~MathIsFun286
 +
 +
==Solution 6==
 +
By change of base formula,
 +
<cmath>
 +
\frac{\log_{2x} 22x}{\log_{2x} 20x} = \frac{{\log_{2x} 11} + 1}{{\log_{2x} 10} + 1} = {\log_{2x} 101} + 1
 +
</cmath>
 +
<cmath>
 +
\log_{2x} 11 + 1 = (\log_{2x} 10)(\log_{2x} 101) + \log{2x} 1010 + 1
 +
</cmath>
 +
<cmath>
 +
\frac{\log_{2x} \frac{11}{1010}}{\log_{2x} 10} = \log_{2x} 101
 +
</cmath>
 +
<cmath>
 +
\log_{10} {\frac{11}{1010}} = \log_{2x} 101
 +
</cmath>
 +
<cmath>
 +
\log_{10} {\frac{11}{1010}} + 1 = \log_{2x} 101 + 1 = \log_{2x} 202x = \log_{20x} {22x}
 +
</cmath>
 +
Thus,
 +
<cmath>
 +
\log_{20x} 22x = \log_{10} \left( \frac{11}{1010} \times 10 \right) = \log_{10} \frac{11}{101}
 +
</cmath>
 +
The requested answer is <math>11 + 101 = \boxed{112}</math>.
 +
 +
~ adam_zheng
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 17:14, 30 January 2024

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

Solution 6

By change of base formula, \[\frac{\log_{2x} 22x}{\log_{2x} 20x} = \frac{{\log_{2x} 11} + 1}{{\log_{2x} 10} + 1} = {\log_{2x} 101} + 1\] \[\log_{2x} 11 + 1 = (\log_{2x} 10)(\log_{2x} 101) + \log{2x} 1010 + 1\] \[\frac{\log_{2x} \frac{11}{1010}}{\log_{2x} 10} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} + 1 = \log_{2x} 101 + 1 = \log_{2x} 202x = \log_{20x} {22x}\] Thus, \[\log_{20x} 22x = \log_{10} \left( \frac{11}{1010} \times 10 \right) = \log_{10} \frac{11}{101}\] The requested answer is $11 + 101 = \boxed{112}$.

~ adam_zheng

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 (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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