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Difference between revisions of "1959 AHSME Problems/Problem 11"

(Add problem statement)
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== Solution ==
 
== Solution ==
−
<cmath>\log_{2}{0.625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.</cmath>
+
<cmath>\log_{2}{0.0625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.</cmath>

Revision as of 12:21, 21 July 2024

Problem

The logarithm of $.0625$ to the base $2$ is: $\textbf{(A)}\ .025 \qquad\textbf{(B)}\ .25\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ -4\qquad\textbf{(E)}\ -2$

Solution

\[\log_{2}{0.0625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.\]

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