Difference between revisions of "2020 AMC 12B Problems/Problem 13"
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==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}>2.5</math> and <math>\log_3{6}>1.5.</math> Therefore, <cmath>\sqrt{\log_2{6}+\log_3{6}}>\sqrt{2.5+1.5}=2.</cmath> Only choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math> are greater than <math>2,</math> but <math>\textbf{(E)}</math> is certainly incorrect--if we compare the squares of the original expression and <math>\textbf{(E)},</math> then they are clearly not equal. So, the answer is <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.</math | + | Using the knowledge of the powers of <math>2</math> and <math>3,</math> we know that <math>\log_2{6}>2.5</math> and <math>\log_3{6}>1.5.</math> Therefore, <cmath>\sqrt{\log_2{6}+\log_3{6}}>\sqrt{2.5+1.5}=2.</cmath> Only choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math> are greater than <math>2,</math> but <math>\textbf{(E)}</math> is certainly incorrect--if we compare the squares of the original expression and <math>\textbf{(E)},</math> then they are clearly not equal. So, the answer is <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.</math> |
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~Baolan | ~Baolan | ||
Revision as of 14:02, 11 July 2021
Contents
Problem
Which of the following is the value of 
Solution 1 (Logic)
Using the knowledge of the powers of 
and 
we know that 
and 
Therefore, ![\[\sqrt{\log_2{6}+\log_3{6}}>\sqrt{2.5+1.5}=2.\]](http://latex.artofproblemsolving.com/a/1/8/a18f439786905da9ecd7595969c89a991c263013.png)
Only choices 
and 
are greater than 
but is certainly incorrect--if we compare the squares of the original expression and 
then they are clearly not equal. So, the answer is 
~Baolan
~Solasky (first edit on wiki!)
~chrisdiamond10
~MRENTHUSIASM (reformatted and merged the thoughts of all contributors)
Solution 2
. If we call 
, then we have
. So our answer is 
.
~JHawk0224
Solution 3
![\[\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\frac{\log{6}}{\log{2}} + \frac{\log{6}}{\log{3}}} = \sqrt{\frac{\log{6}\cdot\log{3} + \log{6}\cdot\log{2}}{\log{3}\cdot\log{2}}} = \sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}}.\]](http://latex.artofproblemsolving.com/f/0/9/f0949839ce59d0e6cdb1ad0577b924b8ed87e940.png)
From here,
![\[\sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2 + \log 3)(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}}.\]](http://latex.artofproblemsolving.com/8/3/5/835715ae9daf1eec876a3750518e0c2e81c5b00e.png)
Finally,
![\[\sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}} = \sqrt{\frac{(\log2 + \log3)^2}{\log 2\cdot\log 3}} = \frac{\log 2}{\sqrt{\log 2\cdot\log 3}} + \frac{\log 3}{\sqrt{\log 2\cdot\log 3}} = \sqrt{\frac{\log 2}{\log 3}} + \sqrt{\frac{\log 3}{\log 2}} = \sqrt{\log_3 2} + \sqrt{\log_2 3}.\]](http://latex.artofproblemsolving.com/3/0/e/30e2b22af48425b666c6bbdcb5811b99079d04da.png)
Answer: 
~ TheBeast5520
Note that in this solution, even the most minor steps have been written out. In the actual test, this solution would be quite fast, and much of it could easily be done in your head.
Video Solution
~IceMatrix
Video Solution
https://youtu.be/RdIIEhsbZKw?t=1463
~ pi_is_3.14
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1298
~ pi_is_3.14
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. 
