1959 AHSME Problems/Problem 11

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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1959 AHSME Problems/Problem 11

Problem

Solution