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

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:

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

$\log_b a>c$ 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

@@ Line 4: / Line 4: @@
 <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>?
+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 ==
 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;">
@@ Line 21: / Line 21: @@
 &\implies x>2001^{2002},
 \end{align*}</cmath>
-from which <math>c=\boxed{\text {(B) }2001^{2002}}.</math>
+from which <math>c=\boxed{\textbf {(B) }2001^{2002}}.</math>
-~Azjps (Fundamental Logic)
+~Azjps ~MRENTHUSIASM
-~MRENTHUSIASM (Reconstruction)
+== 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 ==

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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