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Difference between revisions of "2020 AMC 12B Problems/Problem 13"

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== Solutions ==
 
== Solutions ==
 
=== Solution 1 (Logic) ===
 
=== Solution 1 (Logic) ===
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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> 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 E, 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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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> 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> 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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Written collectively:
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Collaborators:
  
 
~Baolan  
 
~Baolan  

Revision as of 22:55, 14 February 2021

Problem

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

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

Solutions

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 $1.5.$ Only choices $\textbf{(D)}$ and $\textbf{(E)}$ are greater than $2,$ but $\textbf{(E)}$ is certainly incorrect--if we compare the squares of the original expression and $\textbf{(E)},$ they are clearly not equal. So, the answer is $\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.$

Collaborators:

~Baolan

~Solasky (first edit on wiki!)

~chrisdiamond10

~MRENTHUSIASM

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

Solution 3

\[\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\frac{\log{6}}{\log{2}} + \frac{\log{6}}{\log{3}}} = \sqrt{\frac{\log{6}\cdot\log{3} + \log{6}\cdot\log{2}}{\log{3}\cdot\log{2}}} = \sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}}.\] From here, \[\sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2 + \log 3)(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}}.\] Finally, \[\sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}} = \sqrt{\frac{(\log2 + \log3)^2}{\log 2\cdot\log 3}} = \frac{\log 2}{\sqrt{\log 2\cdot\log 3}} + \frac{\log 3}{\sqrt{\log 2\cdot\log 3}} = \sqrt{\frac{\log 2}{\log 3}} + \sqrt{\frac{\log 3}{\log 2}} = \sqrt{\log_3 2} + \sqrt{\log_2 3}.\] Answer: $\boxed{\textbf{(D) } \sqrt{\log_2{3}} + \sqrt{\log_3{2}}}$ ~ 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

https://youtu.be/0xgTR3UEqbQ

~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

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