Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Undo revision 162599 by MRENTHUSIASM (talk))
(Tag: Undo)
(Combined two solutions so they are less repetitive. Also, cleaned up the page a bit. CREDITS ARE RETAINED.)
Line 4: Line 4:
 
<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 (Properties of Logarithms: Direct) ==
+
== Solution 1 (Properties of Logarithms) ==
 
Note that:
 
Note that:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
Line 17: Line 17:
 
&=\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
+
~MRENTHUSIASM (Solution)
  
−
== Solution 2 (Properties of Logarithms: Stepwise) ==
+
~JHawk0224 (Proposal)
−
<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) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}</math>.
+
== Solution 2 (Change of Base Formula)==
−
 
 
−
~JHawk0224
 
−
 
 
−
== Solution 3 (Change of Base Formula)==
 
 
First,
 
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>
Line 44: Line 39:
 
~ TheBeast5520
 
~ TheBeast5520
  
−
==Solution 4 (Observations)==
+
==Solution 3 (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>  
  
Line 55: Line 50:
 
~MRENTHUSIASM (reformatted and merged the thoughts of all contributors)
 
~MRENTHUSIASM (reformatted and merged the thoughts of all contributors)
  
−
== Solution 5 (Solution 4 but More Detailed)==
+
== Solution 4 (Solution 3 but More Detailed)==
 
Note: Only use this method if all else fails and you cannot find a way to simplify the logarithms.
 
Note: Only use this method if all else fails and you cannot find a way to simplify the logarithms.
  

Revision as of 07:38, 22 September 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 (Properties of Logarithms)

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)

~JHawk0224 (Proposal)

Solution 2 (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. On the actual test, this solution would be quite fast, and much of it could easily be done in your head.

~ TheBeast5520

Solution 3 (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 4 (Solution 3 but More Detailed)

Note: Only use this method if all else fails and you cannot find a way to simplify the logarithms.

We can see that $\log_2{6}$ is greater than $2$ and less than $3.$ Additionally, since $6$ is halfway between $2^2$ and $2^3,$ knowing how exponents increase more the larger $x$ is, we can deduce that $\log_2{6}$ is just above halfway between $2$ and $3.$ We can guesstimate this as $\log_2{6} \approx 2.55.$ (It's actually about $2.585.$)

Next, we think of $\log_3{6}.$ This is greater than $1$ and less than $2.$ As $6$ is halfway between $3^1$ and $3^2,$ and similar to the logic for $\log_2{6},$ we know that $\log_3{6}$ is just above halfway between $1$ and $2.$ We guesstimate this as $\log_3{6} \approx 1.55.$ (It's actually about $1.631.$)

So, $\log_2{6} + \log_3{6}$ is approximately $4.1.$ The square root of that is just above $2,$ maybe $2.02.$ We cross out all choices below $\textbf{(C)}$ since they are less than $2,$ and $\textbf{(E)}$ can't possibly be true unless either $\log_2{6}$ and/or $\log_3{6}$ is $0$ (You can prove this by squaring.). Thus, the only feasible answer is $\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.$

-PureSwag

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_13&oldid=162601"