Difference between revisions of "2020 AMC 12B Problems/Problem 13"
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<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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+ | ==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 | ||
==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 |
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