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Difference between revisions of "1962 AHSME Problems/Problem 4"

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So our answer is <math>\boxed{\textbf{(B)}\ \frac{5}{3}}</math>
 
So our answer is <math>\boxed{\textbf{(B)}\ \frac{5}{3}}</math>
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==See Also==
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 21:24, 10 November 2013

Problem

If $8^x = 32$, then x equals:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ \frac{5}{3}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{1}{4}$

Solution

Recognizing that $8=2^3$, we know that $2^{3x}=32$. Since $2^5=32$, we have $2^{3x}=2^5$. Therefore, $x=\dfrac{5}{3}$.

So our answer is $\boxed{\textbf{(B)}\ \frac{5}{3}}$

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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