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Difference between revisions of "2004 AMC 12A Problems/Problem 16"

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<math>\textbf {(A) } 0\qquad \textbf {(B) }2001^{2002} \qquad \textbf {(C) }2002^{2003} \qquad \textbf {(D) }2003^{2004} \qquad \textbf {(E) }2001^{2002^{2003}}</math>
 
<math>\textbf {(A) } 0\qquad \textbf {(B) }2001^{2002} \qquad \textbf {(C) }2002^{2003} \qquad \textbf {(D) }2003^{2004} \qquad \textbf {(E) }2001^{2002^{2003}}</math>
  
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== Solution ==
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== Solution 1 ==
 
For all real numbers <math>a,b,</math> and <math>c</math> such that <math>b>1,</math> note that:
 
For all real numbers <math>a,b,</math> and <math>c</math> such that <math>b>1,</math> note that:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
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from which <math>c=\boxed{\textbf {(B) }2001^{2002}}.</math>
 
from which <math>c=\boxed{\textbf {(B) }2001^{2002}}.</math>
  
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~Azjps (Fundamental Logic)
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~Azjps ~MRENTHUSIASM
  
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~MRENTHUSIASM (Reconstruction)
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== Solution 2 ==
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Let
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Another solution
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<cmath>\begin{align*}
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x &= 2001^a, \\
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2001^a=x
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a &= 2002^b, \\
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b &= 2003^c, \\
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2002^b = a
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c &= 2004^d.
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\end{align*}</cmath>
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2003^c = b
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It follows that <cmath>x = 2001^{2002^{2003^{2004^d}}}.</cmath>
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The smallest value of <math>x</math> occurs when <math>d\rightarrow -\infty,</math> so this expression becomes
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2004^d = c
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<cmath>x = 2001^{2002^{2003^0}} = 2001^{2002^1} = \boxed{\textbf {(B) }2001^{2002}}.</cmath>
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we can now rewrite the expression as x = 2001^(2002^(2003^(2004^d)))
 
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the smallest value of x occurs when d approaches negative infinity. This makes the expression become
 
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1.2001^(2002^(2003^0))
 
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2.2001^(2002^1)
 
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which gives a final expression of 2001^2002
 
  
 
==Video Solution (Logical Thinking)==
 
==Video Solution (Logical Thinking)==

Latest revision as of 01:01, 23 January 2023

Problem

The set of all real numbers $x$ for which

\[\log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x})))\]

is defined is $\{x\mid x > c\}$. What is the value of $c$?

$\textbf {(A) } 0\qquad \textbf {(B) }2001^{2002} \qquad \textbf {(C) }2002^{2003} \qquad \textbf {(D) }2003^{2004} \qquad \textbf {(E) }2001^{2002^{2003}}$

Solution 1

For all real numbers $a,b,$ and $c$ such that $b>1,$ note that:

  1. $\log_b a$ is defined if and only if $a>0.$

  2. $\log_b a>c$ if and only if $a>b^c.$

Therefore, we have \begin{align*} \log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x}))) \text{ is defined} &\implies \log_{2003}(\log_{2002}(\log_{2001}{x}))>0 \\ &\implies \log_{2002}(\log_{2001}{x})>1 \\ &\implies \log_{2001}{x}>2002 \\ &\implies x>2001^{2002}, \end{align*} from which $c=\boxed{\textbf {(B) }2001^{2002}}.$

~Azjps ~MRENTHUSIASM

Solution 2

Let \begin{align*} x &= 2001^a, \\ a &= 2002^b, \\ b &= 2003^c, \\ c &= 2004^d. \end{align*} It follows that \[x = 2001^{2002^{2003^{2004^d}}}.\] The smallest value of $x$ occurs when $d\rightarrow -\infty,$ so this expression becomes \[x = 2001^{2002^{2003^0}} = 2001^{2002^1} = \boxed{\textbf {(B) }2001^{2002}}.\]

Video Solution (Logical Thinking)

https://youtu.be/46c-VN1QzWk

~Education, the Study of Everything

See also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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