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

(Added in Sol 5.)
m (→‎Solution 5: Added in the nonzero condition.)
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<cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath>
 
<cmath>\log_{p^n}{q^n}=\log_{p}{q},</cmath>
 
which can be proven by the Change of Base Formula: <cmath>\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.</cmath>
 
which can be proven by the Change of Base Formula: <cmath>\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.</cmath>
−
Now, we rewrite the original equation as follows:
+
Note that <math>\log_{16}{n}\neq2,</math> so we rewrite the original equation as follows:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\log_4{(\log_{16}{n})^2}&=\log_4{(\log_4{n})} \\
 
\log_4{(\log_{16}{n})^2}&=\log_4{(\log_4{n})} \\

Revision as of 16:29, 19 May 2021

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 1

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 this would lead to use taking the log of $0$), 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 $2+5+6=\boxed{\textbf{(E) }13}.$

Solution 5

We will apply the following property of logarithms: \[\log_{p^n}{q^n}=\log_{p}{q},\] which can be proven by the Change of Base Formula: \[\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.\] Note that $\log_{16}{n}\neq2,$ so we rewrite the original equation as follows: \begin{align*} \log_4{(\log_{16}{n})^2}&=\log_4{(\log_4{n})} \\ (\log_{16}{n})^2&=\log_4{n} \\ (\log_{16}{n})^2&=\log_{16}{n^2} \\ (\log_{16}{n})^2&=2\log_{16}{n} \\ \log_{16}{n}&=2, \end{align*} from which $n=16^2=256.$ The sum of its digits is $2+5+6=\boxed{\textbf{(E) } 13}.$

~MRENTHUSIASM

Video Solution 1

https://youtu.be/fzZzGqNqW6U

~IceMatrix

Video Solution 2

https://youtu.be/RdIIEhsbZKw?t=814

~ pi_is_3.14

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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