Difference between revisions of "2020 AMC 12A Problems/Problem 10"

(Solution)
(Solution)
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becomes
 
becomes
  
<cmath>\log_2{\frac{1}{4}\log_{2}{n}} = \frac{1}{2}\log_2({\frac{1}{2}\log_2{n}}).</cmath>
+
<cmath>\log_2({\frac{1}{4}\log_{2}{n}}) = \frac{1}{2}\log_2({\frac{1}{2}\log_2{n}}).</cmath>
 +
 
 +
Using <math>\log</math> property of addition, we can expand the parentheses into
 +
 
 +
<cmath>\log_2{(\frac{1}{4})}+\log_2{(\log_{2}{n}}) = \frac{1}{2}(\log_2{(\frac{1}{2})} +\log_{2}{(\log_2{n})}).</cmath>

Revision as of 10:32, 1 February 2020

Solution

Any logarithm in the form $\log_{a^b} c = \frac{1}{b} \log_a c$.

so \[\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}.\]

becomes

\[\log_2({\frac{1}{4}\log_{2}{n}}) = \frac{1}{2}\log_2({\frac{1}{2}\log_2{n}}).\]

Using $\log$ property of addition, we can expand the parentheses into

\[\log_2{(\frac{1}{4})}+\log_2{(\log_{2}{n}}) = \frac{1}{2}(\log_2{(\frac{1}{2})} +\log_{2}{(\log_2{n})}).\]