Difference between revisions of "2014 AMC 12A Problems/Problem 18"
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\textbf{(E) }511\qquad</math> | \textbf{(E) }511\qquad</math> | ||
− | ==Solution== | + | ==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>. | ||
− | The domain of <math>\log_{\frac{1}{2}}x</math> is <math>x \in (0, \infty)</math>, so <math> | + | The domain of <math>\log_{\frac{1}{2}}x</math> is <math>x \in (0, \infty)</math>, so <math>d \in (0, \infty)</math>. |
Thus, <math>\log_4{c} \in (0, \infty) \Rightarrow c \in (1, \infty)</math>. | Thus, <math>\log_4{c} \in (0, \infty) \Rightarrow c \in (1, \infty)</math>. | ||
Since <math>c=\log_{\frac{1}{4}}b</math> we have <math>b \in \left(0, \left(\frac{1}{4}\right)^1\right)=\left(0, \frac{1}{4}\right)</math>. | Since <math>c=\log_{\frac{1}{4}}b</math> we have <math>b \in \left(0, \left(\frac{1}{4}\right)^1\right)=\left(0, \frac{1}{4}\right)</math>. | ||
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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{ | + | 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 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. | ||
− | Therefore, the range of <math>f^{-1}(x)</math> is the open interval <math>\left(\lim_{x\to\infty}f(x), \lim_{x\to-\infty}f(x)\right)</math>. We find: | + | Therefore, the range of <math>f^{-1}(x)</math> is the open interval <math>\left(\lim_{x\to\infty}f^{-1}(x), \lim_{x\to-\infty}f^{-1}(x)\right)</math>. We find: |
<cmath>\begin{align*} | <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_{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}}}\\ | ||
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&= \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}}}\\ | |
&= \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 domain of <math>f(x)</math> is <math>\left(\frac{1}{256},\frac{1}{16}\right)</math> and the answer is <math>\boxed{ | + | 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 19:04, 10 July 2021
Contents
Problem
The domain of the function is an interval of length
, where
and
are relatively prime positive integers. What is
?
Solution 1 (Generalization)
For all real numbers and
such that
and
note that:
is defined if and only if
- For
we conclude that:
if and only if
if and only if
For
we conclude that:
if and only if
if and only if
Therefore, we have
The domain of
is an interval of length
from which the answer is
Remark
This problem is quite similar to 2004 AMC 12A Problem 16.
~MRENTHUSIASM
Solution 2 (Substitution)
For simplicity, let , and
.
The domain of is
, so
.
Thus,
.
Since
we have
.
Since
, we have
.
Finally, since
,
.
The length of the interval is
and the answer is
.
Solution 3 (Calculus)
The domain of is the range of the inverse function
. Now
can be seen to be strictly decreasing, since
is decreasing, so
is decreasing, so
is increasing, so
is increasing, therefore
is decreasing.
Therefore, the range of is the open interval
. We find:
Similarly,
Hence the range of
(which is then the domain of
) is
and the answer is
.
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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