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

Revision as of 10:35, 1 February 2020

There is a unique positive integer $n$ such that $\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.$ What is the sum of the digits of $n?$

$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

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 4: / Line 4: @@
 <math>\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13</math>
-[[2020 AMC 12A Problems/Problem 10|Solution]]
 ==Solution==