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

Revision as of 11:42, 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

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

Expanding the RHS and simplifying the logs without variables, we have

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

Subtracting $\frac{1}{2}(\log_{2}{(\log_2{n})})$ from both sides and adding $2$ to both sides gives us

$\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}$

Multiplying by $2$ , raising the logs to exponents of base $2$ to get rid of the logs and simplifying gives us

$\log_{2}{(\log_2{n})} = 3$

$\cancel{2^{\log_{2}}{(\log_2{n})}} = 2^3$

@@ Line 9: / Line 9: @@
 Any logarithm in the form <math>\log_{a^b} c = \frac{1}{b} \log_a c</math>.
-so <cmath>\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}.</cmath>
+so <cmath>\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}</cmath>
 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>
+<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>
 Expanding the RHS and simplifying the logs without variables, we have
-<cmath>-2+\log_2{(\log_{2}{n}}) = -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})}).</cmath>
+<cmath>-2+\log_2{(\log_{2}{n}}) = -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})})</cmath>
+Subtracting <math>\frac{1}{2}(\log_{2}{(\log_2{n})})</math> from both sides and adding <math>2</math> to both sides gives us
+<cmath>\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}</cmath>
+Multiplying by <math>2</math>, raising the logs to exponents of base <math>2</math> to get rid of the logs and simplifying gives us
+<cmath>\log_{2}{(\log_2{n})} = 3</cmath>
+<cmath>\cancel{2^{\log_{2}}{(\log_2{n})}} = 2^3</cmath>