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

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<math>\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13</math>
 
<math>\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13</math>
  
==Solution==
+
==Solution 1==
  
 
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>).
 
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>).
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~ciceronii
 
~ciceronii
  
 
+
==Solution 3 (Change of Base)==
==Solution 3-Change of Base==
 
 
Using the change of base formula on the RHS of the initial equation yields  
 
Using the change of base formula on the RHS of the initial equation yields  
 
<cmath> \log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}} </cmath>
 
<cmath> \log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}} </cmath>
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Canceling out the logs gives
 
Canceling out the logs gives
 
<cmath> (\log_{16}{n})^2=\log_4{n} </cmath>
 
<cmath> (\log_{16}{n})^2=\log_4{n} </cmath>
We use change of base on the RHS to see that <cmath> (log_{16}{n})^2=\frac{ log_{16}{n}}{\log_{16}{4}} </cmath> or <cmath> (log_{16}{n})^2= 2 log_{16}{n}</cmath> Substituting in <math> m = log_{16}{n} </math> gives <math> m^2=2m </math>, so <math> m </math> is either <math>0</math> or <math>2</math>. Since <math> m=0 </math> yields no solution for <math>n</math>(since a log cannot be equal to <math>0</math>), we get <math> 2 = log_{16}{n} </math>, or <math> n=16^2=256 </math>, for a sum of <math>2 + 5 + 6 = \boxed{\textbf{(E) }13}</math>. ~aop2014
+
We use change of base on the RHS to see that <cmath> (\log_{16}{n})^2=\frac{ \log_{16}{n}}{\log_{16}{4}} </cmath> or <cmath> (\log_{16}{n})^2= 2 \log_{16}{n}</cmath> Substituting in <math> m = \log_{16}{n} </math> gives <math> m^2=2m </math>, so <math> m </math> is either <math>0</math> or <math>2</math>. Since <math> m=0 </math> yields no solution for <math>n</math>(since this would lead to use taking the log of <math>0</math>), we get <math> 2 = \log_{16}{n} </math>, or <math> n=16^2=256 </math>, for a sum of <math>2 + 5 + 6 = \boxed{\textbf{(E) }13}</math>. ~aop2014
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 +
==Solution 4==
 +
Suppose <math>\log_2(\log_{16}n)=k\implies\log_{16}n=2^k\implies n=16^{2^k}.</math> Similarly, we have <math>\log_4(\log_4 n)=k\implies \log_4 n=4^k\implies n=4^{4^k}.</math> Thus, we have <cmath>16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}</cmath> and <cmath>4^{4^k}=4^{2^{2k}},</cmath> so <math>k+1=2k\implies k=1.</math> Plugging this in to either one of the expressions for <math>n</math> gives <math>256</math>, and the requested answer is <math>2+5+6=\boxed{\textbf{(E) }13}.</math>
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 +
==Solution 5==
 +
We will apply the following property of logarithms:
 +
<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>
 +
Note that <math>\log_{16}{n}\neq0,</math> so we rewrite the original equation as follows:
 +
<cmath>\begin{align*}
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\log_4{(\log_{16}{n})^2}&=\log_4{(\log_4{n})} \\
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(\log_{16}{n})^2&=\log_4{n} \\
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(\log_{16}{n})^2&=\log_{16}{n^2} \\
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(\log_{16}{n})^2&=2\log_{16}{n} \\
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\log_{16}{n}&=2,
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\end{align*}</cmath>
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from which <math>n=16^2=256.</math> The sum of its digits is <math>2+5+6=\boxed{\textbf{(E) } 13}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Video Solution 1==
 +
https://youtu.be/fzZzGqNqW6U
 +
 
 +
~IceMatrix
 +
 
 +
== Video Solution 2==
 +
https://youtu.be/RdIIEhsbZKw?t=814
 +
 
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 17:31, 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 $\log_{16}{n}\neq0,$ 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 (ProblemsAnswer KeyResources)
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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