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 8: / Line 8: @@
 <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 23: / Line 23: @@
 from which <math>c=\boxed{\textbf {(B) }2001^{2002}}.</math>
-~Azjps (Fundamental Logic)
+~Azjps ~MRENTHUSIASM
-~MRENTHUSIASM (Reconstruction)
+== Solution 2 ==
+Let
-Another solution (I did not know the proper methodology for doing this)
+<cmath>\begin{align*}
+x &= 2001^a, \\
-^a=x
+a &= 2002^b, \\
+b &= 2003^c, \\
-^b = a
+c &= 2004^d.
+\end{align*}</cmath>
-^c = b
+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
-^d = c
+<cmath>x = 2001^{2002^{2003^0}} = 2001^{2002^1} = \boxed{\textbf {(B) }2001^{2002}}.</cmath>
-we can now rewrite the expression as x = 2001^(2002^(2003^(2004^d)))
-the smallest value of x occurs when d approaches negative infinity. This makes the expression become
-.2001^(2002^(2003^0))
-.2001^(2002^1)
-which gives a final expression of 2001^2002
 ==Video Solution (Logical Thinking)==

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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