Difference between revisions of "1959 AHSME Problems/Problem 11"
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| − | ==Solution== | + | == Problem == |
| − | <cmath>\log_{2}{0. | + | |
| + | The logarithm of <math>.0625</math> to the base <math>2</math> is: | ||
| + | <math>\textbf{(A)}\ .025 \qquad\textbf{(B)}\ .25\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ -4\qquad\textbf{(E)}\ -2 </math> | ||
| + | |||
| + | == Solution == | ||
| + | <cmath>\log_{2}{0.0625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.</cmath> | ||
| + | |||
| + | ==See also== | ||
| + | {{AHSME 50p box|year=1959|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:22, 21 July 2024
Problem
The logarithm of 
to the base 
is:
Solution
![\[\log_{2}{0.0625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.\]](http://latex.artofproblemsolving.com/c/a/5/ca5d91d9f245fa7918fe7284f5cdb6f9be7d64a5.png)
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
