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Difference between revisions of "1965 AHSME Problems/Problem 3"

(Created page with "== Problem == The expression <math>(81)^{-2^{-2}}</math> has the same value as: <math>\textbf{(A)}\ \frac {1}{81} \qquad \textbf{() }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \...")
 
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The expression <math>(81)^{-2^{-2}}</math> has the same value as:
 
The expression <math>(81)^{-2^{-2}}</math> has the same value as:
  
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<math>\textbf{(A)}\ \frac {1}{81} \qquad \textbf{() }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4</math>
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<math>\textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B) }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4</math>
  
 
== Solution ==
 
== Solution ==
  
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We know that <math>81^{-2^{-2}}</math> is equivalent to <math>81^{\frac{1}{-2^2}}=81^{\frac{1}{4}}</math>, which is the same as <math>\sqrt[4]{81}=\boxed{3}</math>.
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Let us recall <math>\text{PEMDAS}</math>. We  calculate the exponent first. <math>(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}</math> When we substitute, we get <math>81^{\frac{1}{4}}=\sqrt[4]{81}=\boxed{\textbf{(C) }3}</math>.
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~Mathfun1000
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==See Also==
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{{AHSME box|year=1965|num-b=2|num-a=4}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 20:15, 10 January 2023

Problem

The expression $(81)^{-2^{-2}}$ has the same value as:

$\textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B) }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4$

Solution

Let us recall $\text{PEMDAS}$. We calculate the exponent first. $(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}$ When we substitute, we get $81^{\frac{1}{4}}=\sqrt[4]{81}=\boxed{\textbf{(C) }3}$.

~Mathfun1000

See Also

1965 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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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