https://artofproblemsolving.com/wiki/index.php?title=1979_AHSME_Problems/Problem_9&feed=atom&action=history1979 AHSME Problems/Problem 9 - Revision history2024-03-28T21:53:05ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1979_AHSME_Problems/Problem_9&diff=82191&oldid=prevE power pi times i: Created page with "== Problem 9 == The product of <math>\sqrt[3]{4}</math> and <math>\sqrt[4]{8}</math> equals <math>\textbf{(A) }\sqrt[7]{12}\qquad \textbf{(B) }2\sqrt[7]{12}\qquad \textbf{..."2017-01-05T17:17:14Z<p>Created page with "== Problem 9 == The product of <math>\sqrt[3]{4}</math> and <math>\sqrt[4]{8}</math> equals <math>\textbf{(A) }\sqrt[7]{12}\qquad \textbf{(B) }2\sqrt[7]{12}\qquad \textbf{..."</p>
<p><b>New page</b></p><div>== Problem 9 ==<br />
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The product of <math>\sqrt[3]{4}</math> and <math>\sqrt[4]{8}</math> equals<br />
<br />
<math>\textbf{(A) }\sqrt[7]{12}\qquad<br />
\textbf{(B) }2\sqrt[7]{12}\qquad<br />
\textbf{(C) }\sqrt[7]{32}\qquad<br />
\textbf{(D) }\sqrt[12]{32}\qquad<br />
\textbf{(E) }2\sqrt[12]{32} </math> <br />
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==Solution==<br />
Solution by e_power_pi_times_i<br />
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<math>\sqrt[3]{4}</math> and <math>\sqrt[4]{8}</math> can be expressed as <math>2^{\frac{2}{3}}</math> and <math>2^{\frac{3}{4}}</math>, so their product is <math>2^{\frac{17}{12}} = \boxed{\textbf{(E) } 2\sqrt[12]{32}}</math>.<br />
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== See also ==<br />
{{AHSME box|year=1979|num-b=8|num-a=10}} <br />
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[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>E power pi times i