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

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==Problem==
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== Problem ==
 
Which of the following is the value of <math>\sqrt{\log_2{6}+\log_3{6}}?</math>
 
Which of the following is the value of <math>\sqrt{\log_2{6}+\log_3{6}}?</math>
  
 
<math>\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}}</math>
 
<math>\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}}</math>
  
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== Solutions ==
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=== 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 <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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==Solution 1 (Logic)==
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Collaborators:
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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>. 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
 
  
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Actually, this solution is incomplete, as <math>\sqrt{\log_2{6}} + \sqrt{\log_3{6}}</math> is also greater than 2.    ~chrisdiamond10
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~Baolan
  
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==Solution 2==
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~Solasky (first edit on wiki!)
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~chrisdiamond10
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~MRENTHUSIASM (merged the thoughts of four people)
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=== Solution 2 ===
 
<math>\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}}</math>. If we call <math>\log_2{3} = x</math>, then we have
 
<math>\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}}</math>. If we call <math>\log_2{3} = x</math>, then we have
  
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~JHawk0224
 
~JHawk0224
  
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Answer Choice E is also greater than <math>2,</math> but it’s obvious that it’s too big.
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=== Solution 3 ===
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<cmath>\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}}.</cmath>
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From here,
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<cmath>\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}}.</cmath>
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Finally,
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<cmath>\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}.</cmath>
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Answer: <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}} + \sqrt{\log_3{2}}}</math>
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~ TheBeast5520
  
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~Solasky (first edit on wiki!)
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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.
  
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==Video Solution==
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== Video Solution ==
 
https://youtu.be/0xgTR3UEqbQ
 
https://youtu.be/0xgTR3UEqbQ
  
 
~IceMatrix
 
~IceMatrix
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== Video Solution ==
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https://youtu.be/RdIIEhsbZKw?t=1463
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~ pi_is_3.14
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== Video Solution (Meta-Solving Technique) ==
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https://youtu.be/GmUWIXXf_uk?t=1298
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 05:06, 15 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 (merged the thoughts of four people)

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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