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

## Problem

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

## Solution

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 $$(\log_{2}{(\log_2{n})}) = 3$$ $$2^{\log_{2}{(\log_2{n})}} = 2^3$$ $$\log_2{n}=8$$ $$2^{\log_2{n}}=2^8$$ $$n=256$$

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

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