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

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== Problem ==
 
== Problem ==
−
The [[set]] of all [[real number]]s <math>x</math> for which
+
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>
  
−
is defined as <math>\{x|x > c\}</math>. What is the value of <math>c</math>?
+
is defined is <math>\{x\mid x > c\}</math>. What is the value of <math>c</math>?
  
−
<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>
+
<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>
  
−
== Solution ==
+
== Solution 1 ==
−
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>.
+
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;">
 +
  <li><math>\log_b a</math> is defined if and only if <math>a>0.</math></li><p>
 +
  <li><math>\log_b a>c</math> if and only if <math>a>b^c.</math></li><p>
 +
</ol>
 +
Therefore, we have
 +
<cmath>\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*}</cmath>
 +
from which <math>c=\boxed{\textbf {(B) }2001^{2002}}.</math>
 +
 
 +
~Azjps ~MRENTHUSIASM
 +
 
 +
== Solution 2 ==
 +
Let
 +
<cmath>\begin{align*}
 +
x &= 2001^a, \\
 +
a &= 2002^b, \\
 +
b &= 2003^c, \\
 +
c &= 2004^d.
 +
\end{align*}</cmath>
 +
It follows that <cmath>x = 2001^{2002^{2003^{2004^d}}}.</cmath>
 +
The smallest value of <math>x</math> occurs when <math>d\rightarrow -\infty,</math> so this expression becomes
 +
<cmath>x = 2001^{2002^{2003^0}} = 2001^{2002^1} = \boxed{\textbf {(B) }2001^{2002}}.</cmath>
 +
 
 +
==Video Solution (Logical Thinking)==
 +
https://youtu.be/46c-VN1QzWk
 +
 
 +
~Education, the Study of Everything
  
 
== See also ==
 
== See also ==
−
{{AMC12 box|year=2004|ab=A|num-b=14|num-a=16}}
+
{{AMC12 box|year=2004|ab=A|num-b=15|num-a=17}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

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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