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

Revision as of 10:48, 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 $\log_{a^b} c = \frac{1}{b} \log_a c$ . (this can be proved easily by using change of base formula to base $a$ ).

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

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

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

$\log_2{n}=8$

$2^{\log_2{n}}=2^8$

$n=256$

Adding the digits together, we have $2+5+6=\boxed{\textbf{E) }13}$ ~quacker88

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 12 Problems and Solutions

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 ==Solution==
-Any logarithm in the form <math>\log_{a^b} c = \frac{1}{b} \log_a c</math>.
+Any logarithm in the form <math>\log_{a^b} c = \frac{1}{b} \log_a c</math>. (this can be proved easily by using change of base formula to base <math>a</math>).
 so <cmath>\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}</cmath>