Difference between revisions of "2020 AMC 12B Problems/Problem 13"
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&=\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}. | &=\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | ~ | + | ~HUASHILIAO2020 (Solution) |
~JHawk0224 (Proposal) | ~JHawk0224 (Proposal) |
Revision as of 17:34, 5 December 2021
Contents
Problem
Which of the following is the value of
Solution 1 (Properties of Logarithms)
Recall that:
We use these properties of logarithms to rewrite the original expression: ~HUASHILIAO2020 (Solution)
~JHawk0224 (Proposal)
Solution 2 (Change of Base Formula)
First, From here, Finally, Answer:
Note that in this solution, even the most minor steps have been written out. On the actual test, this solution would be quite fast, and much of it could easily be done in your head.
~ TheBeast5520
Solution 3 (Observations)
Using the knowledge of the powers of and we know that and Therefore, Only choices and are greater than but is certainly incorrect--if we compare the squares of the original expression and then they are clearly not equal. So, the answer is
~Baolan
~Solasky (first edit on wiki!)
~chrisdiamond10
~MRENTHUSIASM (reformatted and merged the thoughts of all contributors)
Solution 4 (Solution 3 but More Detailed)
Note: Only use this method if all else fails and you cannot find a way to simplify the logarithms.
We can see that is greater than and less than Additionally, since is halfway between and knowing how exponents increase more the larger is, we can deduce that is just above halfway between and We can guesstimate this as (It's actually about )
Next, we think of This is greater than and less than As is halfway between and and similar to the logic for we know that is just above halfway between and We guesstimate this as (It's actually about )
So, is approximately The square root of that is just above maybe We cross out all choices below since they are less than and can't possibly be true unless either and/or is (You can prove this by squaring.). Thus, the only feasible answer is
-PureSwag
Video Solution
~IceMatrix
Video Solution
https://youtu.be/RdIIEhsbZKw?t=1463
~ pi_is_3.14
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1298
~ pi_is_3.14
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.