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Difference between revisions of "2020 AMC 12B Problems/Problem 13"

(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.)
(→‎Solution 2 (Properties of Logarithms: Direct))
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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
  
 
== Solution 3 (Properties of Logarithms: Stepwise) ==
 
== Solution 3 (Properties of Logarithms: Stepwise) ==

Revision as of 01:02, 12 July 2021

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 (Observations)

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 (Properties of Logarithms: Direct)

Note that:

  1. $\log_b{(uv)}=\log_b u + \log_b v.$

  2. $\log_b u\cdot\log_u b=1.$

We use these properties of logarithms to rewrite the original expression: \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*} ~MRENTHUSIASM

Solution 3 (Properties of Logarithms: Stepwise)

$\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\log_2{2}+\log_2{3}+\log_3{2}+\log_3{3}}=\sqrt{2+\log_2{3}+\log_3{2}}$. 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 $\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}$.

~JHawk0224

Solution 4 (Change of Base Formula)

First, \[\sqrt{\log_2{6}+\log_3{6}} = \sqrt{\frac{\log{6}}{\log{2}} + \frac{\log{6}}{\log{3}}} = \sqrt{\frac{\log{6}\cdot\log{3} + \log{6}\cdot\log{2}}{\log{3}\cdot\log{2}}} = \sqrt{\frac{\log{6}(\log 2 + \log 3)}{\log 2\cdot \log 3}}.\] 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, \begin{align*} \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*} Answer: $\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}$

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

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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