2004 AMC 12A Problems/Problem 16

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

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2004 AMC 12A Problems/Problem 16

Contents

Problem

Solution 1

Solution 2

Video Solution (Logical Thinking)

See also