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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 (Properties of Logarithms)==
  
−
Any logarithm in the form <math>\log_{a^b} c = \frac{1}{b} \log_a c</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 <cmath>\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}</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>
 +
Using log property of addition, we expand both sides and then simplify:
 +
<cmath>\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*}</cmath>
 +
Subtracting <math>\frac{1}{2}(\log_{2}{(\log_2{n})})</math> from both sides and adding <math>2</math> to both sides gives us <cmath>\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}.</cmath>
 +
Multiplying by <math>2,</math> exponentiating, and simplifying gives us
 +
<cmath>\begin{align*}
 +
\log_{2}{(\log_2{n})} &= 3 \\
 +
\log_2{n}&=8 \\
 +
n&=256.
 +
\end{align*}</cmath>
 +
Adding the digits together, we have <math>2+5+6=\boxed{\textbf{(E) } 13}.</math>  
  
−
becomes
+
~quacker88 (Solution)
  
−
<cmath>\log_2({\frac{1}{4}\log_{2}{n}}) = \frac{1}{2}\log_2({\frac{1}{2}\log_2{n}})</cmath>
+
~MRENTHUSIASM (Reformatting)
  
−
Using <math>\log</math> property of addition, we can expand the parentheses into
+
==Solution 2 (Properties of Logarithms)==
 +
We will apply the following logarithmic identity:
 +
<cmath>\log_{p^k}{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:
 +
<cmath>\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*}</cmath>
 +
from which <math>n=16^2=256.</math> The sum of its digits is <math>2+5+6=\boxed{\textbf{(E) } 13}.</math>
  
−
<cmath>\log_2{(\frac{1}{4})}+\log_2{(\log_{2}{n}}) = \frac{1}{2}(\log_2{(\frac{1}{2})} +\log_{2}{(\log_2{n})})</cmath>
+
~MRENTHUSIASM
  
−
Expanding the RHS and simplifying the logs without variables, we have
+
==Solution 3 (Properties of Logarithms)==
 +
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>
 +
This means we can multiply each side by <math>2</math> for
 +
<cmath> \log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})}. </cmath>
 +
Canceling out the logs gives
 +
<cmath> (\log_{16}{n})^2=\log_4{n}. </cmath>
 +
We use change of base on the RHS to see that
 +
<cmath>\begin{align*}
 +
(\log_{16}{n})^2&=\frac{ \log_{16}{n}}{\log_{16}{4}} \\
 +
(\log_{16}{n})^2&=2 \log_{16}{n}.
 +
\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>
  
−
<cmath>-2+\log_2{(\log_{2}{n}}) = -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})})</cmath>
+
~aop2014
  
−
Subtracting <math>\frac{1}{2}(\log_{2}{(\log_2{n})})</math> from both sides and adding <math>2</math> to both sides gives us
+
==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>
  
−
<cmath>\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}</cmath>
+
==Solution 5 (Guess and Check)==
 +
We know that, as the answer is an integer, <math>n</math> must be some power of <math>16.</math> Testing <math>16</math> yields
 +
<cmath>\begin{align*}
 +
\log_2{(\log_{16}{16})} &= \log_4{(\log_4{16})} \\
 +
\log_2{1} &= \log_4{2} \\
 +
0 &= \frac{1}{2},
 +
\end{align*}</cmath>
 +
which does not work. We then try <math>256,</math> giving us
 +
<cmath>\begin{align*}
 +
\log_2{(\log_{16}{256})} &= \log_4{(\log_4{256})} \\
 +
\log_2{2} &= \log_4{4} \\
 +
1 &= 1,
 +
\end{align*}</cmath>
 +
which holds true. Thus, <math>n = 256,</math> so the answer is <math>2 + 5 + 6 = \boxed{\textbf{(E) } 13}.</math>
  
−
Multiplying by <math>2</math>, raising the logs to exponents of base <math>2</math> to get rid of the logs and simplifying gives us
+
(Don't use this technique unless you absolutely need to! Guess and check methods aren't helpful for learning math.)
  
−
<cmath>\log_{2}{(\log_2{n})} = 3</cmath>
+
~ciceronii (Solution)
  
−
<cmath>\cancel{2^{\log_{2}}}{(\log_2{n})}} = 2^3</cmath>
+
~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==
 +
 
 +
{{AMC12 box|year=2020|ab=A|num-b=9|num-a=11}}
 +
{{MAA Notice}}

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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