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

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== Problem ==
 
== Problem ==
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The [[set]] of all [[real number]]s <math>x</math> for which
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The set of all real numbers <math>x</math> for which
  
 
<cmath>\log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x})))</cmath>
 
<cmath>\log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x})))</cmath>
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is defined is <math>\{x|x > c\}</math>. What is the value of <math>c</math>?
 
is defined is <math>\{x|x > c\}</math>. What is the value of <math>c</math>?
  
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<math>\text {(A)} 0\qquad \text {(B)}2001^{2002} \qquad \text {(C)}2002^{2003} \qquad \text {(D)}2003^{2004} \qquad \text {(E)}2001^{2002^{2003}}</math>
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<math>\text {(A) } 0\qquad \text {(B) }2001^{2002} \qquad \text {(C) }2002^{2003} \qquad \text {(D) }2003^{2004} \qquad \text {(E) }2001^{2002^{2003}}</math>
  
 
== Solution ==
 
== Solution ==
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We know that the domain of <math>\log_k n</math>, where <math>k</math> is a [[constant]], is <math>n > 0</math>. So <math>\log_{2003}(\log_{2002}(\log_{2001}{x})) > 0</math>. By the definition of [[logarithm]]s, we then have <math>\log_{2002}(\log_{2001}{x})) > 2003^0 = 1</math>. Then <math>\log_{2001}{x} > 2002^1 = 2002</math> and <math>x > 2001^{2002}\ \mathrm{(B)}</math>.
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For all real numbers <math>b</math> such that <math>b>0</math> and <math>b\neq1,</math> note that:
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<ol style="margin-left: 1.5em;">
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  <li><math>\log_b a</math> is defined if and only if <math>a>0.</math></li><p>
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  <li><math>\log_b a>c</math> if and only if <math>a>b^c.</math></li><p>
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</ol>
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Therefore, we have
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<cmath>\begin{align*}
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\log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x}))) \text{ is defined} &\implies \log_{2003}(\log_{2002}(\log_{2001}{x}))>0 \\
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&\implies \log_{2002}(\log_{2001}{x})>1 \\
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&\implies \log_{2001}{x}>2002 \\
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&\implies \boxed{\text {(B) }2001^{2002}}.
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\end{align*}</cmath>
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~Azjps (Fundamental Logic)
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 +
~MRENTHUSIASM (Reconstruction)
  
 
== See also ==
 
== See also ==

Revision as of 01:50, 10 July 2021

Problem

The set of all real numbers $x$ for which

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

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

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

Solution

For all real numbers $b$ such that $b>0$ and $b\neq1,$ 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 \boxed{\text {(B) }2001^{2002}}. \end{align*} ~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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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