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1965 AHSME Problems/Problem 3: Difference between revisions

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== Solution ==
== Solution ==


Let us recall <math>\text{PEMDAS}</math>. We realize that we have to calculate the exponent first. <math>(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}</math> When we substitute, we get <math>81^{\dfrac{1}{4}}=\sqrt[4]{81}=\boxed{\textbf{(C) }3}</math>.
Let us recall <math>\text{PEMDAS}</math>. We realize that we have to 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>.


~Mathfun1000 (Explaining clearly)
~Mathfun1000 (Explaining clearly)

Revision as of 08:07, 18 August 2021

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 realize that we have to 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 (Explaining clearly)

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