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

Revision as of 21:12, 3 October 2014

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

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

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 $\boxed{\textbf{(B)}\ \frac{5}{3}}$

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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 ==See Also==
-{{AHSME 50p box|year=1962|before=Problem 3|num-a=5}}
+{{AHSME 40p box|year=1962|before=Problem 3|num-a=5}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}