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1962 AHSME Problems/Problem 4: Difference between revisions

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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>
==See Also==
[[Category:Introductory Algebra Problems]]
{{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

These problems are copyrighted © by the Mathematical Association of America.

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