Difference between revisions of "2020 AMC 12B Problems/Problem 13"
m (→Video Solution) |
(→Solution 2) |
||
| Line 11: | Line 11: | ||
<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 | ||
| − | <math>\sqrt{2+x+\frac{1}{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{\log_2{3}}+\frac{1}{\sqrt{\ | + | <math>\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}}</math>. So our answer is <math>\boxed{\textbf{(D)}}</math>. |
~JHawk0224 | ~JHawk0224 | ||
Revision as of 23:00, 8 March 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
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. 
