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2020 AMC 12B Problems/Problem 13

Revision as of 15:10, 11 July 2021 by MRENTHUSIASM (talk | contribs) (Undo revision 157932 by MRENTHUSIASM (talk))

Problem

Which of the following is the value of $\sqrt{\log_2{6}+\log_3{6}}?$

$\textbf{(A) } 1 \qquad\textbf{(B) } \sqrt{\log_5{6}} \qquad\textbf{(C) } 2 \qquad\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}} \qquad\textbf{(E) } \sqrt{\log_2{6}}+\sqrt{\log_3{6}}$

Solution 1 (Logic)

Using the knowledge of the powers of $2$ and $3,$ we know that $\log_2{6}>2.5$ and $\log_3{6}>1.5.$ Therefore, \[\sqrt{\log_2{6}+\log_3{6}}>\sqrt{2.5+1.5}=2.\] Only choices $\textbf{(D)}$ and $\textbf{(E)}$ are greater than $2,$ but $\textbf{(E)}$ is certainly incorrect--if we compare the squares of the original expression and $\textbf{(E)},$ then they are clearly not equal. So, the answer is $\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.$

~Baolan

~Solasky (first edit on wiki!)

~chrisdiamond10

~MRENTHUSIASM (reformatted and merged the thoughts of all contributors)

Solution 2

$\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\log_2{2}+\log_2{3}+\log_3{2}+\log_3{3}}=\sqrt{2+\log_2{3}+\log_3{2}}$. If we call $\log_2{3} = x$, then we have

$\sqrt{2+x+\frac{1}{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{\log_2{3}}+\frac{1}{\sqrt{\log_2{3}}}=\sqrt{\log_2{3}}+\sqrt{\log_3{2}}$. So our answer is $\boxed{\textbf{(D)}}$.

~JHawk0224

Solution 3

\[\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\frac{\log{6}}{\log{2}} + \frac{\log{6}}{\log{3}}} = \sqrt{\frac{\log{6}\cdot\log{3} + \log{6}\cdot\log{2}}{\log{3}\cdot\log{2}}} = \sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}}.\] From here, \[\sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2 + \log 3)(\log 2 + \log 3)}{\log 2\cdot \log 3}} = \sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}}.\] Finally, \[\sqrt{\frac{(\log 2)^2 + 2\cdot\log2\cdot\log3 + (\log3)^2}{\log 2\cdot\log 3}} = \sqrt{\frac{(\log2 + \log3)^2}{\log 2\cdot\log 3}} = \frac{\log 2}{\sqrt{\log 2\cdot\log 3}} + \frac{\log 3}{\sqrt{\log 2\cdot\log 3}} = \sqrt{\frac{\log 2}{\log 3}} + \sqrt{\frac{\log 3}{\log 2}} = \sqrt{\log_3 2} + \sqrt{\log_2 3}.\] Answer: $\boxed{\textbf{(D) } \sqrt{\log_2{3}} + \sqrt{\log_3{2}}}$ ~ TheBeast5520

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.

Video Solution

https://youtu.be/0xgTR3UEqbQ

~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

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