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 5)
(→‎Solution 1)
Line 7: Line 7:
 
==Solution 1==
 
==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>
  
−
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 can expand the parentheses into <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].</cmath>
 +
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>
 +
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
 +
<math></math>\begin{align*}
 +
(\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.
 +
\end{align*}
 +
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
 
−
 
 
−
<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>
 
−
 
 
−
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>
 
−
 
 
−
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>, raising the logs to exponents of base <math>2</math> to get rid of the logs and simplifying gives us
 
−
 
 
−
<cmath>(\log_{2}{(\log_2{n})}) = 3</cmath>
 
−
 
 
−
<cmath>2^{\log_{2}{(\log_2{n})}} = 2^3</cmath>
 
−
 
 
−
<cmath>\log_2{n}=8</cmath>
 
−
 
 
−
<cmath>2^{\log_2{n}}=2^8</cmath>
 
−
 
 
−
<cmath>n=256</cmath>
 
−
 
 
−
Adding the digits together, we have <math>2+5+6=\boxed{\textbf{(E) }13}</math> ~quacker88
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 07:06, 15 August 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, 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 can expand the parentheses into \[\log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) = \frac{1}{2}\left[\log_2{\frac{1}{2}} +\log_{2}{(\log_2{n})}\right].\] 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 $2,$ exponentiating, and simplifying gives us $$ (Error compiling LaTeX. Unknown error_msg)\begin{align*} (\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. \end{align*} Adding the digits together, we have $2+5+6=\boxed{\textbf{(E) }13}.$

~quacker88 (Solution)

~MRENTHUSIASM (Reformatting)

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 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

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

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=160294"