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 ==
==Solution 1 (Logic)==
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=== 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.
 
Answer Choice E is also greater than <math>2,</math> but it’s obvious that it’s too big.
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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
 
Actually, this solution is incomplete, as <math>\sqrt{\log_2{6}} + \sqrt{\log_3{6}}</math> is also greater than 2.    ~chrisdiamond10
  
==Solution 2==
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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
  
==Video Solution==
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== Video Solution ==
 
https://youtu.be/0xgTR3UEqbQ
 
https://youtu.be/0xgTR3UEqbQ
  

Revision as of 13:29, 19 January 2021

Problem

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

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

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 $1.5$. 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!)

Specifically, verify Choice E is too big by squaring the expression in the question and squaring choice (E) and then comparing.

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 (ProblemsAnswer KeyResources)
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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