Difference between revisions of "2020 AMC 12B Problems/Problem 13"
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| − | ==Problem== | + | == 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> | ||
| − | + | == Solutions == | |
| − | ==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 | ||
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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== | + | === 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== | + | == 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 
Solutions
Solution 1 (Logic)
Using the knowledge of the powers of 
and 
, we know that 
is greater than 
and 
is greater than 
. So that means 
. Since 
is the only option greater than , it's the answer. ~Baolan
Answer Choice E is also greater than 
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 
is also greater than 2. ~chrisdiamond10
Solution 2
. If we call 
, then we have
. So our answer is 
.
~JHawk0224
Video Solution
~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. 
