Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
(→‎Solution 2 (Properties of Logarithms))
(12 intermediate revisions by 4 users not shown)
Line 5: Line 5:
 
<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]] 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>
−
Using log property of addition, we can expand the LHS into <cmath>\log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}).</cmath>
+
Using log property of addition, we expand both sides and then simplify:
−
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>
+
<cmath>\begin{align*}
−
Expanding the RHS and simplifying the logs without variables, we have <cmath>-2+\log_2{(\log_{2}{n}}) = -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})}).</cmath>
+
\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>
 
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
 
Multiplying by <math>2,</math> exponentiating, and simplifying gives us
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
(\log_{2}{(\log_2{n})}) &= 3 \\
+
\log_{2}{(\log_2{n})} &= 3 \\
−
2^{\log_{2}{(\log_2{n})}} &= 2^3 \\
 
 
\log_2{n}&=8 \\
 
\log_2{n}&=8 \\
−
2^{\log_2{n}}&=2^8 \\
 
 
n&=256.
 
n&=256.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
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)
Line 28: Line 29:
 
~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*}
Line 44: Line 45:
 
~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>
Line 56: Line 57:
 
(\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)==
Line 76: Line 77:
 
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.)
Line 84: Line 85:
 
~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==

Revision as of 01:32, 30 January 2023

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_10&oldid=188375"