1959 AHSME Problems/Problem 11: Difference between revisions
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== Solution == | == Solution == | ||
<cmath>\log_{2}{0. | <cmath>\log_{2}{0.0625}=\log_{2}{\frac{1}{16}}=\boxed{-4}.</cmath> | ||
Revision as of 11:21, 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)
