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

m (→‎Solution 1: Deleted some intermediate steps, as they are unnecessary to readers. Also, kept the spaces cleaner.)
(→‎Solution 1 (Properties of Logarithms))
 
(12 intermediate revisions by 4 users not shown)
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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 1==
+
==Solution 1 (Properties of Logarithms)==
  
−
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>
+
We can use the fact that <math>\log_{a^b} c = \frac{1}{b} \log_a c.</math> This can be proved by using [[change of base formula | Change of Base Formula]] to base <math>a.</math>
  
 
So, the original equation <math>\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}</math> becomes <cmath>\log_2\left({\frac{1}{4}\log_{2}{n}}\right) = \frac{1}{2}\log_2\left({\frac{1}{2}\log_2{n}}\right).</cmath>
 
So, the original equation <math>\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}</math> becomes <cmath>\log_2\left({\frac{1}{4}\log_{2}{n}}\right) = \frac{1}{2}\log_2\left({\frac{1}{2}\log_2{n}}\right).</cmath>
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n&=256.
 
n&=256.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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Adding the digits together, we have <math>2+5+6=\boxed{\textbf{(E) }13}.</math>  
+
Adding the digits together, we have <math>2+5+6=\boxed{\textbf{(E) } 13}.</math>  
  
 
~quacker88 (Solution)
 
~quacker88 (Solution)
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~MRENTHUSIASM (Reformatting)
 
~MRENTHUSIASM (Reformatting)
  
−
==Solution 2==
+
==Solution 2 (Properties of Logarithms)==
 
We will apply the following logarithmic identity:
 
We will apply the following logarithmic identity:
 
<cmath>\log_{p^k}{q^k}=\log_{p}{q},</cmath>
 
<cmath>\log_{p^k}{q^k}=\log_{p}{q},</cmath>
−
which can be proven by the Change of Base Formula: <cmath>\log_{p^k}{q^k}=\frac{\log_{p}{q^k}}{\log_{p}{p^k}}=\frac{k\log_{p}{q}}{k}=\log_{p}{q}.</cmath>
+
which can be proven by the [[change of base formula | Change of Base Formula]]: <cmath>\log_{p^k}{q^k}=\frac{\log_{p}{q^k}}{\log_{p}{p^k}}=\frac{k\log_{p}{q}}{k}=\log_{p}{q}.</cmath>
 
Note that <math>\log_{16}{n}\neq0,</math> so we rewrite the original equation as follows:
 
Note that <math>\log_{16}{n}\neq0,</math> so we rewrite the original equation as follows:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 3 (Change of Base)==
+
==Solution 3 (Properties of Logarithms)==
 
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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(\log_{16}{n})^2&=2 \log_{16}{n}.
 
(\log_{16}{n})^2&=2 \log_{16}{n}.
 
\end{align*}</cmath>
 
\end{align*}</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 the digit-sum of <math>2 + 5 + 6 = \boxed{\textbf{(E) }13}.</math>
+
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 the digit-sum of <math>2 + 5 + 6 = \boxed{\textbf{(E) } 13}.</math>
  
 
~aop2014
 
~aop2014
  
 
==Solution 4 (Exponential Form)==
 
==Solution 4 (Exponential Form)==
−
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>
+
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>
  
 
==Solution 5 (Guess and Check)==
 
==Solution 5 (Guess and Check)==
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1 &= 1,
 
1 &= 1,
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
which holds true. Thus, <math>n = 256,</math> so the answer is <math>2 + 5 + 6 = \boxed{\textbf{(E) }13}.</math>
+
which holds true. Thus, <math>n = 256,</math> so the answer is <math>2 + 5 + 6 = \boxed{\textbf{(E) } 13}.</math>
  
 
(Don't use this technique unless you absolutely need to! Guess and check methods aren't helpful for learning math.)
 
(Don't use this technique unless you absolutely need to! Guess and check methods aren't helpful for learning math.)
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~MRENTHUSIASM (Reformatting)
 
~MRENTHUSIASM (Reformatting)
  
−
==Video Solution 1==
+
==Video Solution==
 
https://youtu.be/fzZzGqNqW6U
 
https://youtu.be/fzZzGqNqW6U
  
 
~IceMatrix
 
~IceMatrix
  
−
== Video Solution 2==
+
== Video Solution by OmegaLearn ==
 
https://youtu.be/RdIIEhsbZKw?t=814
 
https://youtu.be/RdIIEhsbZKw?t=814
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution==
 +
https://youtu.be/EnyzIHcJ8Aw
 +
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

Latest revision as of 10:30, 12 September 2024

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 (Properties of Logarithms)

We can use the fact that $\log_{a^b} c = \frac{1}{b} \log_a c.$ This can be proved by using Change of Base Formula to base $a.$

So, the original equation $\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}$ becomes \[\log_2\left({\frac{1}{4}\log_{2}{n}}\right) = \frac{1}{2}\log_2\left({\frac{1}{2}\log_2{n}}\right).\] Using log property of addition, we expand both sides and then simplify: \begin{align*} \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[\log_2{\frac{1}{2}} +\log_{2}{(\log_2{n})}\right] \\ \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[-1 +\log_{2}{(\log_2{n})}\right] \\ -2+\log_2{(\log_{2}{n}}) &= -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})}). \end{align*} 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 $2,$ exponentiating, and simplifying gives us \begin{align*} \log_{2}{(\log_2{n})} &= 3 \\ \log_2{n}&=8 \\ n&=256. \end{align*} Adding the digits together, we have $2+5+6=\boxed{\textbf{(E) } 13}.$

~quacker88 (Solution)

~MRENTHUSIASM (Reformatting)

Solution 2 (Properties of Logarithms)

We will apply the following logarithmic identity: \[\log_{p^k}{q^k}=\log_{p}{q},\] which can be proven by the Change of Base Formula: \[\log_{p^k}{q^k}=\frac{\log_{p}{q^k}}{\log_{p}{p^k}}=\frac{k\log_{p}{q}}{k}=\log_{p}{q}.\] 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

Solution 3 (Properties of Logarithms)

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 \begin{align*} (\log_{16}{n})^2&=\frac{ \log_{16}{n}}{\log_{16}{4}} \\ (\log_{16}{n})^2&=2 \log_{16}{n}. \end{align*} 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 the digit-sum of $2 + 5 + 6 = \boxed{\textbf{(E) } 13}.$

~aop2014

Solution 4 (Exponential Form)

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 (Guess and Check)

We know that, as the answer is an integer, $n$ must be some power of $16.$ Testing $16$ yields \begin{align*} \log_2{(\log_{16}{16})} &= \log_4{(\log_4{16})} \\ \log_2{1} &= \log_4{2} \\ 0 &= \frac{1}{2}, \end{align*} which does not work. We then try $256,$ giving us \begin{align*} \log_2{(\log_{16}{256})} &= \log_4{(\log_4{256})} \\ \log_2{2} &= \log_4{4} \\ 1 &= 1, \end{align*} 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)

~MRENTHUSIASM (Reformatting)

Video Solution

https://youtu.be/fzZzGqNqW6U

~IceMatrix

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/EnyzIHcJ8Aw

~Education, the Study of Everything

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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