1962 AHSME Problems/Problem 4

Revision as of 12:21, 21 February 2017 by Hiabc (talk | contribs) (Problem)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

1962 AHSC (ProblemsAnswer KeyResources)
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

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