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 12B Problems/Problem 13"

m (→‎Solution 2)
(→‎Solution 1 (Logic))
Line 7: Line 7:
 
==Solution 1 (Logic)==
 
==Solution 1 (Logic)==
 
Using the knowledge of the powers of <math>2</math> and <math>3</math>, we know that <math>\log_2{6}</math> is greater than <math>2.5</math> and <math>\log_3{6}</math> is greater than <math>1.5</math>. So that means <math>\sqrt{\log_2{6}+\log_3{6}} > 2</math>. Since <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}} + \sqrt{\log_3{2}}}</math> is the only option greater than <math>2</math>, it's the answer.  ~Baolan
 
Using the knowledge of the powers of <math>2</math> and <math>3</math>, we know that <math>\log_2{6}</math> is greater than <math>2.5</math> and <math>\log_3{6}</math> is greater than <math>1.5</math>. So that means <math>\sqrt{\log_2{6}+\log_3{6}} > 2</math>. Since <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}} + \sqrt{\log_3{2}}}</math> is the only option greater than <math>2</math>, it's the answer.  ~Baolan
 +
 +
 +
Answer Choice E is also greater than <math>2,</math> but it’s obvious that it’s too big.
 +
 +
~Solasky (first edit on wiki!)
  
 
Actually, this solution is incomplete, as <math>\sqrt{\log_2{6}} + \sqrt{\log_3{6}}</math> is also greater than 2.    ~chrisdiamond10
 
Actually, this solution is incomplete, as <math>\sqrt{\log_2{6}} + \sqrt{\log_3{6}}</math> is also greater than 2.    ~chrisdiamond10

Revision as of 16:28, 4 January 2021

Problem

Which of the following is the value of $\sqrt{\log_2{6}+\log_3{6}}?$

$\textbf{(A) } 1 \qquad\textbf{(B) } \sqrt{\log_5{6}} \qquad\textbf{(C) } 2 \qquad\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}} \qquad\textbf{(E) } \sqrt{\log_2{6}}+\sqrt{\log_3{6}}$


Solution 1 (Logic)

Using the knowledge of the powers of $2$ and $3$, we know that $\log_2{6}$ is greater than $2.5$ and $\log_3{6}$ is greater than $1.5$. So that means $\sqrt{\log_2{6}+\log_3{6}} > 2$. Since $\boxed{\textbf{(D) } \sqrt{\log_2{3}} + \sqrt{\log_3{2}}}$ is the only option greater than $2$, it's the answer. ~Baolan


Answer Choice E is also greater than $2,$ but it’s obvious that it’s too big.

~Solasky (first edit on wiki!)

Actually, this solution is incomplete, as $\sqrt{\log_2{6}} + \sqrt{\log_3{6}}$ is also greater than 2. ~chrisdiamond10

Solution 2

$\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\log_2{2}+\log_2{3}+\log_3{2}+\log_3{3}}=\sqrt{2+\log_2{3}+\log_3{2}}$. If we call $\log_2{3} = x$, then we have

$\sqrt{2+x+\frac{1}{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{\log_2{3}}+\frac{1}{\sqrt{\log_2{3}}}=\sqrt{\log_2{3}}+\sqrt{\log_3{2}}$. So our answer is $\boxed{\textbf{(D)}}$.

~JHawk0224

Video Solution

https://youtu.be/0xgTR3UEqbQ

~IceMatrix

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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_12B_Problems/Problem_13&oldid=141507"