Difference between revisions of "2020 AMC 12B Problems/Problem 13"
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MRENTHUSIASM (talk | contribs) (Combined two solutions so they are less repetitive. Also, cleaned up the page a bit. CREDITS ARE RETAINED.) |
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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> | ||
− | == Solution 1 (Properties of Logarithms | + | == Solution 1 (Properties of Logarithms) == |
Note that: | Note that: | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
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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> | ||
− | ~MRENTHUSIASM | + | ~MRENTHUSIASM (Solution) |
− | + | ~JHawk0224 (Proposal) | |
− | |||
− | + | == Solution 2 (Change of Base Formula)== | |
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First, | First, | ||
<cmath>\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\frac{\log{6}}{\log{2}} + \frac{\log{6}}{\log{3}}} = \sqrt{\frac{\log{6}\cdot\log{3} + \log{6}\cdot\log{2}}{\log{3}\cdot\log{2}}} = \sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}}.</cmath> | <cmath>\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\frac{\log{6}}{\log{2}} + \frac{\log{6}}{\log{3}}} = \sqrt{\frac{\log{6}\cdot\log{3} + \log{6}\cdot\log{2}}{\log{3}\cdot\log{2}}} = \sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}}.</cmath> | ||
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~ TheBeast5520 | ~ TheBeast5520 | ||
− | ==Solution | + | ==Solution 3 (Observations)== |
Using the knowledge of the powers of <math>2</math> and <math>3,</math> we know that <math>\log_2{6}>2.5</math> and <math>\log_3{6}>1.5.</math> Therefore, <cmath>\sqrt{\log_2{6}+\log_3{6}}>\sqrt{2.5+1.5}=2.</cmath> Only choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math> are greater than <math>2,</math> but <math>\textbf{(E)}</math> is certainly incorrect--if we compare the squares of the original expression and <math>\textbf{(E)},</math> then they are clearly not equal. So, the answer is <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.</math> | Using the knowledge of the powers of <math>2</math> and <math>3,</math> we know that <math>\log_2{6}>2.5</math> and <math>\log_3{6}>1.5.</math> Therefore, <cmath>\sqrt{\log_2{6}+\log_3{6}}>\sqrt{2.5+1.5}=2.</cmath> Only choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}</math> are greater than <math>2,</math> but <math>\textbf{(E)}</math> is certainly incorrect--if we compare the squares of the original expression and <math>\textbf{(E)},</math> then they are clearly not equal. So, the answer is <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.</math> | ||
− | ~Baolan | + | ~Baolan |
− | ~Solasky (first edit on wiki!) | + | ~Solasky (first edit on wiki!) |
− | ~chrisdiamond10 | + | ~chrisdiamond10 |
~MRENTHUSIASM (reformatted and merged the thoughts of all contributors) | ~MRENTHUSIASM (reformatted and merged the thoughts of all contributors) | ||
− | == Solution | + | == 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. | Note: Only use this method if all else fails and you cannot find a way to simplify the logarithms. | ||
Revision as of 08:33, 22 September 2021
Contents
Problem
Which of the following is the value of
Solution 1 (Properties of Logarithms)
Note that:
We use these properties of logarithms to rewrite the original expression:
~MRENTHUSIASM (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.