Difference between revisions of "1959 AHSME Problems/Problem 11"
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| − | ==Solution== | + | == Problem == |
| + | |||
| + | 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}{0.625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.</cmath> | <cmath>\log_{2}{0.625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.</cmath> | ||
Revision as of 13:58, 16 July 2024
Problem
The logarithm of 
to the base 
is:
Solution
![\[\log_{2}{0.625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.\]](http://latex.artofproblemsolving.com/f/f/a/ffab788b9f294873d08f3df9df51920ad1a67670.png)
