2020 AMC 12A Problems/Problem 10

Revision as of 11:45, 1 February 2020 by Quacker88 (talk | contribs) (Solution)


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


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

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

\[(\log_{2}{(\log_2{n})}) = 3\]

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


\[2^{\log_2{n}=2^8\] (Error compiling LaTeX. ! Missing } inserted.)


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

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