Difference between revisions of "1962 AHSME Problems/Problem 4"
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==Problem== | ==Problem== | ||
| − | If <math>8^x = 32</math>, then x equals: | + | If <math>8^x = 32</math>, then <math>x</math> equals: |
<math> \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} </math> | <math> \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} </math> | ||
==Solution== | ==Solution== | ||
| − | + | Recognizing that <math>8=2^3</math>, we know that <math>2^{3x}=32</math>. Since <math>2^5=32</math>, we have <math>2^{3x}=2^5</math>. Therefore, <math>x=\dfrac{5}{3}</math>. | |
| + | |||
| + | So our answer is <math>\boxed{\textbf{(B)}\ \frac{5}{3}}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|before=Problem 3|num-a=5}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:21, 21 February 2017
Problem
If 
, then 
equals:
Solution
Recognizing that 
, we know that 
. Since 
, we have 
. Therefore, 
.
So our answer is 
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
