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

Revision as of 05:59, 18 May 2016

Problem 12

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

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

Solution

Because $\log_2 \Big(\log_3 (\log_2 x) \Big) = 0$ . That means that $(\log_3 (\log_2 x) =1$ . That means that $\log_2 x=3$ . Therefore, $x=8$ . Since $x=8$ , $x^{-1/2}=\frac{1}{2\sqrt{2}}$ . Since this is none of the answer choices, the answer is $\fbox{\textbf{E} \text{None of these}}$

1983 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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 ==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>
+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}{2\sqrt{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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Difference between revisions of "1983 AHSME Problems/Problem 12"

Revision as of 05:59, 18 May 2016

Problem 12

Solution

See Also