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

Revision as of 16:28, 4 January 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 (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

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

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

Video Solution

https://youtu.be/0xgTR3UEqbQ

~IceMatrix

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.

@@ Line 16: / Line 16: @@
 ~JHawk0224
-Answer Choice E is also greater than <math>2,</math> but it’s obvious that it’s too big.
-~Solasky (first edit on wiki!)
 ==Video Solution==

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