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

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==Solution==
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== Problem ==
<cmath>\log_{2}{0.625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.</cmath>
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The logarithm of <math>.0625</math> to the base <math>2</math> is:
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<math>\textbf{(A)}\ .025 \qquad\textbf{(B)}\ .25\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ -4\qquad\textbf{(E)}\ -2    </math>
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== Solution ==
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<cmath>\log_{2}{0.0625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.</cmath>
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==See also==
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{{AHSME 50p box|year=1959|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 12:22, 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

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

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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