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

(→‎Solution 1 (Logic))
(→‎Solution 1 (Logic))
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For A, B, C, D, only D is greater than 2.
 
For A, B, C, D, only D is greater than 2.
 
E is certainly incorrect--if we compare the squares of the original expression and E, they are clearly not equal.
 
E is certainly incorrect--if we compare the squares of the original expression and E, they are clearly not equal.
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So, the answer is <math>\boxed{\textbf{(D)}.</math> ~MRENTHUSIASM
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So, the answer is <math>\boxed{\textbf{(D)}}.</math> ~MRENTHUSIASM
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 18:29, 2 February 2021

Problem

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

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

Solutions

Solution 1 (Logic)

Using the knowledge of the powers of $2$ and $3$, we know that $\log_2{6}$ is greater than $2.5$ and $\log_3{6}$ is greater than $1.5$. So that means $\sqrt{\log_2{6}+\log_3{6}} > 2$. Since $\boxed{\textbf{(D) } \sqrt{\log_2{3}} + \sqrt{\log_3{2}}}$ is the only option greater than $2$, it's the answer. ~Baolan

Answer Choice E is also greater than $2,$ but it’s obvious that it’s too big.

~Solasky (first edit on wiki!)

Specifically, verify Choice E is too big by squaring the expression in the question and squaring choice (E) and then comparing.

Actually, this solution is incomplete, as $\sqrt{\log_2{6}} + \sqrt{\log_3{6}}$ is also greater than 2. ~chrisdiamond10

Using the approximations mentioned above, the original expression is greater than 2. For A, B, C, D, only D is greater than 2. E is certainly incorrect--if we compare the squares of the original expression and E, they are clearly not equal. So, the answer is $\boxed{\textbf{(D)}}.$ ~MRENTHUSIASM

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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