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Difference between revisions of "2014 AMC 12A Problems/Problem 18"

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(→‎Solution)
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==Solution==
 
==Solution==
 +
===Solution 1===
 
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 19:
  
 
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{271 \text{ (C)}}</math>.
 +
 +
===Solution 2===
 +
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.
 +
 +
Therefore, the range of <math>f^{-1}(x)</math> is an open interval and can be found by taking the limits as <math>x</math> approaches <math>-\infty</math> and <math>\infty</math>. In fact, the range is <math>\left(\lim_{x\to\infty}f(x), \lim_{x\to-\infty}f(x)\right)</math>. We find:
 +
<cmath>\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*}</cmath>
 +
 +
Similarly,
 +
<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_{b\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)}}\\
 +
&= \left(\frac1{16}\right)^2\\
 +
&= \frac{1}{256}.
 +
\end{align*}</cmath>
 +
Hence 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>.
  
 
==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}}

Revision as of 21:53, 7 February 2014

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

Solution 1

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 $c \in (0, \infty)$. 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 $\boxed{271 \text{ (C)}}$.

Solution 2

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 an open interval and can be found by taking the limits as $x$ approaches $-\infty$ and $\infty$. In fact, the range is $\left(\lim_{x\to\infty}f(x), \lim_{x\to-\infty}f(x)\right)$. 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}}}\\ &= \lim_{b\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)}}\\ &= \left(\frac1{16}\right)^2\\ &= \frac{1}{256}. \end{align*} Hence the domain of $f(x)$ is $\left(\frac{1}{256},\frac{1}{16}\right)$ and the answer is $\boxed{271 \text{ (C)}}$.

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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