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

2020 AMC 12A Problems/Problem 10

Revision as of 13:34, 3 February 2020 by Aop2014 (talk | contribs)

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

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 a log cannot be equal to $0$), we get $2 = log_{16}{n}$, or $n=16^2=256$, for a sum of $2 + 5 + 6 = \boxed{\textbf{(E) }13}$.

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=116751"