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

Revision as of 14:12, 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

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 $\boxed{271 \text{ (C)}}$ .

2014 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
-{{AMC12 box|year=2014|ab=A|before=17|num-a=19}}
+{{AMC12 box|year=2014|ab=A|num-b=17|num-a=19}}
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Difference between revisions of "2014 AMC 12A Problems/Problem 18"

Revision as of 14:12, 7 February 2014

Problem

Solution

See Also