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Difference between revisions of "2014 AMC 12A Problems/Problem 18"
m (→‎Solution 1 (Generalization))
 
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\textbf{(E) }511\qquad</math>
 
\textbf{(E) }511\qquad</math>
  
−
==Solution 1==
+
==Solution 1 (Generalization)==
 +
For all real numbers <math>a,b,</math> and <math>c</math> such that <math>b>0</math> and <math>b\neq1,</math> note that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>\log_b a</math> is defined if and only if <math>a>0.</math></li><p>
 +
  <li>For <math>0<b<1,</math> we conclude that:
 +
<ul style="list-style-type:square;">
 +
<li><math>\log_b a<c</math> if and only if <math>a>b^c.</math></li><p>
 +
<li><math>\log_b a>c</math> if and only if <math>0<a<b^c.</math></li><p>
 +
</ul>
 +
For <math>b>1,</math> we conclude that:
 +
<ul style="list-style-type:square;">
 +
<li><math>\log_b a<c</math> if and only if <math>0<a<b^c.</math></li><p>
 +
<li><math>\log_b a>c</math> if and only if <math>a>b^c.</math></li><p>
 +
</ul>
 +
</ol>
 +
Therefore, we have
 +
<cmath>\begin{align*}
 +
\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\
 +
&\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))>1 \\
 +
&\implies 0<\log_{16}(\log_{\frac1{16}}x)<\frac14 \\
 +
&\implies 1<\log_{\frac1{16}}x<2 \\
 +
&\implies \frac{1}{256}<x<\frac{1}{16}.
 +
\end{align*}</cmath>
 +
The domain of <math>f(x)</math> is an interval of length <math>\frac{1}{16}-\frac{1}{256}=\frac{15}{256},</math> from which the answer is <math>15+256=\boxed{\textbf{(C) }271}.</math>
 +
 
 +
<u><b>Remark</b></u>
 +
 
 +
This problem is quite similar to [[2004_AMC_12A_Problems/Problem_16|2004 AMC 12A Problem 16]].
  
 +
~MRENTHUSIASM
 +
 +
==Solution 2 (Substitution)==
 
For simplicity, let <math>a=\log_{\frac{1}{16}}{x},b=\log_{16}a,c=\log_{\frac{1}{4}}b</math>, and <math>d=\log_4c</math>.
 
For simplicity, let <math>a=\log_{\frac{1}{16}}{x},b=\log_{16}a,c=\log_{\frac{1}{4}}b</math>, and <math>d=\log_4c</math>.
  
Line 18: Line 48:
 
Finally, since <math>a=\log_{\frac{1}{16}}{x}</math>, <math>x \in \left(\left(\frac{1}{16}\right)^2,\left(\frac{1}{16}\right)^1\right)=\left(\frac{1}{256},\frac{1}{16}\right)</math>.
 
Finally, since <math>a=\log_{\frac{1}{16}}{x}</math>, <math>x \in \left(\left(\frac{1}{16}\right)^2,\left(\frac{1}{16}\right)^1\right)=\left(\frac{1}{256},\frac{1}{16}\right)</math>.
  
−
The length of the <math>x</math> interval is <math>\frac{1}{16}-\frac{1}{256}=\frac{15}{256}</math> and the answer is <math>\boxed{271 \text{ (C)}}</math>.
+
The length of the <math>x</math> interval is <math>\frac{1}{16}-\frac{1}{256}=\frac{15}{256}</math> and the answer is <math>\boxed{\textbf{(C) }271}</math>.
  
−
==Solution 2==
+
==Solution 3 (Calculus)==
 
The domain of <math>f(x)</math> is the range of the inverse function <math>f^{-1}(x)=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}</math>. Now <math>f^{-1}(x)</math> can be seen to be strictly decreasing, since <math>\left(\frac12\right)^x</math> is decreasing, so <math>4^{\left(\frac12\right)^x}</math> is decreasing, so <math>\left(\frac14\right)^{4^{\left(\frac12\right)^x}}</math> is increasing, so <math>16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}</math> is increasing, therefore <math>\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}</math> is decreasing.
 
The domain of <math>f(x)</math> is the range of the inverse function <math>f^{-1}(x)=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}</math>. Now <math>f^{-1}(x)</math> can be seen to be strictly decreasing, since <math>\left(\frac12\right)^x</math> is decreasing, so <math>4^{\left(\frac12\right)^x}</math> is decreasing, so <math>\left(\frac14\right)^{4^{\left(\frac12\right)^x}}</math> is increasing, so <math>16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}</math> is increasing, therefore <math>\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}</math> is decreasing.
  
Line 30: Line 60:
 
&= \frac{1}{16}.
 
&= \frac{1}{16}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
 
 
Similarly,
 
Similarly,
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
  \lim_{x\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^0}}}\\
+
\lim_{x\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^0}}}\\
 
&= \left(\frac1{16}\right)^{16^{\frac14}}\\
 
&= \left(\frac1{16}\right)^{16^{\frac14}}\\
 
&= \left(\frac1{16}\right)^2\\
 
&= \left(\frac1{16}\right)^2\\
 
&= \frac{1}{256}.
 
&= \frac{1}{256}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Hence the range of <math>f^{-1}(x)</math> (which is then the domain of <math>f(x)</math>) is <math>\left(\frac{1}{256},\frac{1}{16}\right)</math> and the answer is <math>\boxed{271 \text{ (C)}}</math>.
+
Hence the range of <math>f^{-1}(x)</math> (which is then the domain of <math>f(x)</math>) is <math>\left(\frac{1}{256},\frac{1}{16}\right)</math> and the answer is <math>\boxed{\textbf{(C) }271}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2014|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:04, 10 July 2021

Problem

The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }511\qquad$

Solution 1 (Generalization)

For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\neq1,$ note that:

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

  2. For $0<b<1,$ we conclude that:
    • $\log_b a<c$ if and only if $a>b^c.$

    • $\log_b a>c$ if and only if $0<a<b^c.$

    For $b>1,$ we conclude that:

    • $\log_b a<c$ if and only if $0<a<b^c.$

    • $\log_b a>c$ if and only if $a>b^c.$

Therefore, we have \begin{align*} \log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\ &\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))>1 \\ &\implies 0<\log_{16}(\log_{\frac1{16}}x)<\frac14 \\ &\implies 1<\log_{\frac1{16}}x<2 \\ &\implies \frac{1}{256}<x<\frac{1}{16}. \end{align*} The domain of $f(x)$ is an interval of length $\frac{1}{16}-\frac{1}{256}=\frac{15}{256},$ from which the answer is $15+256=\boxed{\textbf{(C) }271}.$

Remark

This problem is quite similar to 2004 AMC 12A Problem 16.

~MRENTHUSIASM

Solution 2 (Substitution)

For simplicity, let $a=\log_{\frac{1}{16}}{x},b=\log_{16}a,c=\log_{\frac{1}{4}}b$, and $d=\log_4c$.

The domain of $\log_{\frac{1}{2}}x$ is $x \in (0, \infty)$, so $d \in (0, \infty)$. Thus, $\log_4{c} \in (0, \infty) \Rightarrow c \in (1, \infty)$. Since $c=\log_{\frac{1}{4}}b$ we have $b \in \left(0, \left(\frac{1}{4}\right)^1\right)=\left(0, \frac{1}{4}\right)$. Since $b=\log_{16}{a}$, we have $a \in (16^0,16^{1/4})=(1,2)$. Finally, since $a=\log_{\frac{1}{16}}{x}$, $x \in \left(\left(\frac{1}{16}\right)^2,\left(\frac{1}{16}\right)^1\right)=\left(\frac{1}{256},\frac{1}{16}\right)$.

The length of the $x$ interval is $\frac{1}{16}-\frac{1}{256}=\frac{15}{256}$ and the answer is $\boxed{\textbf{(C) }271}$.

Solution 3 (Calculus)

The domain of $f(x)$ is the range of the inverse function $f^{-1}(x)=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}$. Now $f^{-1}(x)$ can be seen to be strictly decreasing, since $\left(\frac12\right)^x$ is decreasing, so $4^{\left(\frac12\right)^x}$ is decreasing, so $\left(\frac14\right)^{4^{\left(\frac12\right)^x}}$ is increasing, so $16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}$ is increasing, therefore $\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}$ is decreasing.

Therefore, the range of $f^{-1}(x)$ is the open interval $\left(\lim_{x\to\infty}f^{-1}(x), \lim_{x\to-\infty}f^{-1}(x)\right)$. We find: \begin{align*} \lim_{x\to-\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&= \lim_{a\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^a}}}\\ &= \lim_{b\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^b}}\\ &= \left(\frac1{16}\right)^{16^0}\\ &= \frac{1}{16}. \end{align*} Similarly, \begin{align*} \lim_{x\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^0}}}\\ &= \left(\frac1{16}\right)^{16^{\frac14}}\\ &= \left(\frac1{16}\right)^2\\ &= \frac{1}{256}. \end{align*} Hence the range of $f^{-1}(x)$ (which is then the domain of $f(x)$) is $\left(\frac{1}{256},\frac{1}{16}\right)$ and the answer is $\boxed{\textbf{(C) }271}$.

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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