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

(Created page with "==Problem 12== If <math>\log_2 \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...")
 
(→‎Solution)
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==Solution==
 
==Solution==
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Because <math>\log_2 \Big(\log_3 (\log_2 x) \Big) = 0</math>. That means that <math>\Big(\log_3 (\log_2 x) \Big=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}{2sqrt{2}}</math>. Since this is none of the answer choices, the answer is <math>\fbox{\textbf{E} \text{None of these}}</math>
 +
 
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1983|num-b=11|num-a=13}}
 
{{AHSME box|year=1983|num-b=11|num-a=13}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 05:58, 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 $x=8$, $x^{-1/2}=\frac{1}{2sqrt{2}}$. Since this is none of the answer choices, the answer is $\fbox{\textbf{E} \text{None of these}}$

See Also

1983 AHSME (Problems • Answer Key • Resources)
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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