Difference between revisions of "2020 AMC 12B Problems/Problem 13"
MRENTHUSIASM (talk | contribs) m (Undo revision 157932 by MRENTHUSIASM (talk)) (Tag: Undo) |
MRENTHUSIASM (talk | contribs) (Titled all solutions. Also, I added in a more direct solution without using any variables. Lastly, I rearranged all solutions from similar ideas. Let me know if anyone is unhappy with this version. Oh, I reformatted Sol 4 as well.) |
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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 ( | + | ==Solution 1 (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> | ||
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~MRENTHUSIASM (reformatted and merged the thoughts of all contributors) | ~MRENTHUSIASM (reformatted and merged the thoughts of all contributors) | ||
− | == Solution 2 == | + | == Solution 2 (Properties of Logarithms: Direct) == |
+ | Note that: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>\log_b{(uv)}=\log_b u + \log_b v.</math></li><p> | ||
+ | <li><math>\log_b u\cdot\log_u b=1.</math></li><p> | ||
+ | </ol> | ||
+ | We use these properties of logarithms to rewrite the original expression: | ||
+ | <cmath>\begin{align*} | ||
+ | \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}} \ | ||
+ | &=\sqrt{\left(\sqrt{\log_2{3}}+\sqrt{\log_3{2}}\right)^2} \ | ||
+ | &=\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | == Solution 3 (Properties of Logarithms: Stepwise) == | ||
<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{\log_2{3}}}=\sqrt{\log_2{3}}+\sqrt{\log_3{2}}</math>. So our answer is <math>\boxed{\textbf{(D)}}</math>. | + | <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) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}</math>. |
~JHawk0224 | ~JHawk0224 | ||
− | == Solution | + | == Solution 4 (Change of Base Formula)== |
+ | 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> | ||
From here, | From here, | ||
<cmath>\sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2 + \log 3)(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}}.</cmath> | <cmath>\sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2 + \log 3)(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}}.</cmath> | ||
Finally, | Finally, | ||
− | <cmath>\sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}} = \sqrt{\frac{(\log2 + \log3)^2}{\log 2\cdot\log 3}} = \frac{\log 2}{\sqrt{\log 2\cdot\log 3}} + \frac{\log 3}{\sqrt{\log 2\cdot\log 3}} = \sqrt{\frac{\log 2}{\log 3}} + \sqrt{\frac{\log 3}{\log 2}} = \sqrt{\log_3 2} + \sqrt{\log_2 3}.</cmath> | + | <cmath>\begin{align*} |
− | Answer: <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}} + \sqrt{\log_3{2}}}</math> | + | \sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}} &= \sqrt{\frac{(\log2 + \log3)^2}{\log 2\cdot\log 3}} \ |
− | + | &= \frac{\log 2}{\sqrt{\log 2\cdot\log 3}} + \frac{\log 3}{\sqrt{\log 2\cdot\log 3}} \ | |
+ | &= \sqrt{\frac{\log 2}{\log 3}} + \sqrt{\frac{\log 3}{\log 2}} \ | ||
+ | &= \sqrt{\log_3 2} + \sqrt{\log_2 3}. | ||
+ | \end{align*}</cmath> | ||
+ | Answer: <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}</math> | ||
Note that in this solution, even the most minor steps have been written out. In the actual test, this solution would be quite fast, and much of it could easily be done in your head. | Note that in this solution, even the most minor steps have been written out. In the actual test, this solution would be quite fast, and much of it could easily be done in your head. | ||
+ | |||
+ | ~ TheBeast5520 | ||
== Video Solution == | == Video Solution == |
Revision as of 01:02, 12 July 2021
Contents
[hide]Problem
Which of the following is the value of
Solution 1 (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 2 (Properties of Logarithms: Direct)
Note that:
We use these properties of logarithms to rewrite the original expression:
Solution 3 (Properties of Logarithms: Stepwise)
. If we call , then we have
. So our answer is .
~JHawk0224
Solution 4 (Change of Base Formula)
First, From here, Finally, Answer:
Note that in this solution, even the most minor steps have been written out. In the actual test, this solution would be quite fast, and much of it could easily be done in your head.
~ TheBeast5520
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.