# 2020 AMC 12A Problems/Problem 10

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 $\log_{a^b} c = \frac{1}{b} \log_a c$. (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

## Solution 2

We know that, as the answer is an integer, $n$ must be some power of $16$. Testing $16$ yields $$\log_2{(\log_{16}{16})} = \log_4{(\log_4{16})}$$ $$\log_2{1} = \log_4{2}$$ $$0 = \frac{1}{2}$$ which does not work. We then try $256$, giving us

$$\log_2{(\log_{16}{256})} = \log_4{(\log_4{256})}$$ $$\log_2{2} = \log_4{4}$$ $$1 = 1$$ which holds true. Thus, $n = 256$, so the answer is $2 + 5 + 6 = \boxed{\textbf{(E) }13}$.

(Don't use this technique unless you absolutely need to! Guess and check methods aren't helpful for learning math.)

~ciceronii

## Solution 3-Change of Base

Using the change of base formula on the RHS of the initial equation yields $$\log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}}$$ This means we can multiply each side by 2 for $$\log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})}$$ Canceling out the logs gives $$(\log_{16}{n})^2=\log_4{n}$$ We use change of base on the RHS to see that $$(log_{16}{n})^2=\frac{ log_{16}{n}}{\log_{16}{4}}$$ or $$(log_{16}{n})^2= 2 log_{16}{n}$$ Substituting in $m = log_{16}{n}$ gives $m^2=2m$, so $m$ is either $0$ or $2$. Since $m=0$ yields no solution for $n$(since a log cannot be equal to $0$), we get $2 = log_{16}{n}$, or $n=16^2=256$, for a sum of $2 + 5 + 6 = \boxed{\textbf{(E) }13}$. ~aop2014

## Solution 4

Suppose $\log_2(\log_{16}n)=k\implies\log_{16}n=2^k\implies n=16^{2^k}.$ Similarly, we have $\log_4(\log_4 n)=k\implies \log_4 n=4^k\implies n=4^{4^k}.$ Thus, we have $$16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}$$ and $$4^{4^k}=4^{2^{2k}},$$ so $k+1=2k\implies k=1.$ Plugging this in to either one of the expressions for $n$ gives $256$, and the requested answer is $2+5+6=\boxed{\textbf{(E) }13}.$

~IceMatrix