Difference between revisions of "1983 AHSME Problems/Problem 12"

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==Problem 12==
 
==Problem 12==
If <math>\log_2 \Big(\log_3 (\log_2 x) \Big) = 0</math>, then <math>x^{-1/2}</math> equals
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If <math>\log_7 \Big(\log_3 (\log_2 x) \Big) = 0</math>, then <math>x^{-1/2}</math> equals
  
<math>\text{(A)} \ \frac{1}{3} \qquad  \text{(B)} \ \frac{1}{2 \sqrt 3} \qquad  \text{(C)}\ \frac{1}{3\sqrt 3}\qquad \text{(D)}\ \frac{1}{\sqrt{42}}\qquad \text{(E)}\ \text{none of these}</math>
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<math>\textbf{(A)} \ \frac{1}{3} \qquad  \textbf{(B)} \ \frac{1}{2 \sqrt 3} \qquad  \textbf{(C)}\ \frac{1}{3\sqrt 3}\qquad \textbf{(D)}\ \frac{1}{\sqrt{42}}\qquad \textbf{(E)}\ \text{none of these}</math>
  
 
==Solution==
 
==Solution==
Because <math>\log_2 \Big(\log_3 (\log_2 x) \Big) = 0</math>. That means that <math>(\log_3 (\log_2 x) =1</math>. That means that <math>\log_2 x=3</math>. Therefore, <math>x=8</math>. Since <math>x=8</math>, <math>x^{-1/2}=\frac{1}{2sqrt{2}}</math>. Since this is none of the answer choices, the answer is <math>\fbox{\textbf{E} \text{None of these}}</math>
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Because <math>\log_7 \Big(\log_3 (\log_2 x) \Big) = 0</math>, we deduce <math>\log_3 (\log_2 x) =1</math>, and thus <math>\log_2 x=3</math>. Therefore, <math>x=8</math>, which means <math>x^{-1/2}=\frac{1}{2\sqrt{2}}</math>. Since this does not match any of the answer choices, the answer is <math>\fbox{{\bf(E)} \text{none of these}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 23:47, 19 February 2019

Problem 12

If $\log_7 \Big(\log_3 (\log_2 x) \Big) = 0$, then $x^{-1/2}$ equals

$\textbf{(A)} \ \frac{1}{3} \qquad  \textbf{(B)} \ \frac{1}{2 \sqrt 3} \qquad  \textbf{(C)}\ \frac{1}{3\sqrt 3}\qquad \textbf{(D)}\ \frac{1}{\sqrt{42}}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Because $\log_7 \Big(\log_3 (\log_2 x) \Big) = 0$, we deduce $\log_3 (\log_2 x) =1$, and thus $\log_2 x=3$. Therefore, $x=8$, which means $x^{-1/2}=\frac{1}{2\sqrt{2}}$. Since this does not match any of the answer choices, the answer is $\fbox{{\bf(E)} \text{none of these}}$.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 26 27 28 29 30
All AHSME Problems and Solutions


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