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

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m (Problem)
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<cmath>\log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x})))</cmath>
 
<cmath>\log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x})))</cmath>
  
is defined is <math>\{x|x > c\}</math>. What is the value of <math>c</math>?
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is defined is <math>\{x\mid x > c\}</math>. What is the value of <math>c</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>
 
<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>

Revision as of 09:01, 6 September 2021

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

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 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
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