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

Revision as of 14:05, 23 June 2016

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

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

We know that $81^{-2^{-2}}$ is equivalent to $81^{\frac{1}{-2^2}}=81^{\frac{1}{4}}$ , which is the same as $\sqrt[4]{81}=\boxed{\textbf{(C) }3}$ .

Revision as of 14:04, 23 June 2016 (view source) Hiabc (talk \| contribs) m (→‎Added (B) choice) ← Older edit		Revision as of 14:05, 23 June 2016 (view source) Hiabc (talk \| contribs) (→‎Solution) Newer edit →
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	== Solution ==		== Solution ==

−	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>.	+	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{\textbf{(C) }3}</math>.