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

Revision as of 10:32, 1 February 2020

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

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

@@ Line 7: / Line 7: @@
 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>