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}</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 |
==See Also== | ==See Also== | ||
Revision as of 21:12, 7 February 2020
Problem
Which of the following is the value of 
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
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. 
