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

Revision as of 23:52, 21 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 $\begin{align*} 2001^a &= x, \\ 2002^b &= a, \\ 2003^c &= b, \\ 2004^d &= c. \end{align*}$ It follows that $x = 2001^{2002^{2003^{2004^d}}}.$ The smallest value of $x$ occurs when $d\rightarrow -\infty.$ This makes the 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 26: / Line 26: @@
 == Solution 2 ==
+Let
-^a=x
+<cmath>\begin{align*}
+^a &= x, \\
-^b = a
+^b &= a, \\
+^c &= b, \\
-^c = b
+^d &= c.
+\end{align*}</cmath>
-^d = c
+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> This makes the expression becomes
-we can now rewrite the expression as x = 2001^(2002^(2003^(2004^d)))
+<cmath>x = 2001^{2002^{2003^0}} = 2001^{2002^1} = \boxed{\textbf {(B) }2001^{2002}}.</cmath>
-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
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
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