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

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== Solution ==
 
== Solution ==
For all real numbers <math>b</math> such that <math>b>0</math> and <math>b\neq1,</math> note that:
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For all real numbers <math>a</math> and <math>b</math> such that <math>b>0</math> and <math>b\neq1,</math> note that:
 
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   <li><math>\log_b a</math> is defined if and only if <math>a>0.</math></li><p>
 
   <li><math>\log_b a</math> is defined if and only if <math>a>0.</math></li><p>

Revision as of 13:07, 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 $a$ and $b$ 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 x>2001^{2002}, \end{align*} from which $c=\boxed{\text {(B) }2001^{2002}}.$

~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
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