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

(Ordered the solutions by elegance. PM me if you disagree with this order--I pushed the observation solutions to the end.)
m (Video Solution (Meta-Solving Technique))
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<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) ==
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== Solution 1 (Properties of Logarithms) ==
Note that:
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Recall that:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li><math>\log_b{(uv)}=\log_b u + \log_b v.</math></li><p>
 
   <li><math>\log_b{(uv)}=\log_b u + \log_b v.</math></li><p>
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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
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~MRENTHUSIASM (Solution)
  
== Solution 2 (Properties of Logarithms: Stepwise) ==
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~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>
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~ TheBeast5520
 
~ TheBeast5520
  
==Solution 4 (Observations)==
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==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>  
  
~Baolan  
+
~Baolan
  
~Solasky (first edit on wiki!)  
+
~Solasky (first edit on wiki!)
  
~chrisdiamond10  
+
~chrisdiamond10
  
 
~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.
  
We can see that <math>\log_2{6}</math> is greater than <math>2</math> and less than <math>3</math>. Additionally, since <math>6</math> is halfway between <math>2^2</math> and <math>2^3</math>, knowing how exponents increase more the larger <math>x</math> is, we can deduce that <math>\log_2{6}</math> is just above halfway between <math>2</math> and <math>3</math>. We can guesstimate this as <math>\log_2{6} \approx 2.55.</math> (It's actually about <math>2.585.</math>)
+
We can see that <math>\log_2{6}</math> is greater than <math>2</math> and less than <math>3.</math> Additionally, since <math>6</math> is halfway between <math>2^2</math> and <math>2^3,</math> knowing how exponents increase more the larger <math>x</math> is, we can deduce that <math>\log_2{6}</math> is just above halfway between <math>2</math> and <math>3.</math> We can guesstimate this as <math>\log_2{6} \approx 2.55.</math> (It's actually about <math>2.585.</math>)
  
Next, we think of <math>\log_3{6}.</math> This is greater than <math>1</math> and less than <math>2</math>. <math>6</math> is halfway between <math>3^1</math> and <math>3^2,</math> and similar to the logic for <math>\log_2{6},</math> we know that <math>\log_3{6}</math> is just above halfway between <math>1</math> and <math>2</math>. We guesstimate this as <math>\log_3{6} \approx 1.55.</math> (It's actually about <math>1.631.</math>)
+
Next, we think of <math>\log_3{6}.</math> This is greater than <math>1</math> and less than <math>2.</math> As <math>6</math> is halfway between <math>3^1</math> and <math>3^2,</math> and similar to the logic for <math>\log_2{6},</math> we know that <math>\log_3{6}</math> is just above halfway between <math>1</math> and <math>2.</math> We guesstimate this as <math>\log_3{6} \approx 1.55.</math> (It's actually about <math>1.631.</math>)
  
So <math>\log_2{6} + \log_3{6}</math> is approximately <math>4.1.</math> The square root of that is just above <math>2,</math> maybe <math>2.02.</math> We cross out all choices below <math>C</math> since they are less than <math>2</math>, and <math>E</math> can't possibly be true unless either <math>\log_2{6}</math> and/or <math>\log_3{6}</math> is <math>0</math> (you can prove this by squaring). Thus, the only feasible answer is <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.</math>
+
So, <math>\log_2{6} + \log_3{6}</math> is approximately <math>4.1.</math> The square root of that is just above <math>2,</math> maybe <math>2.02.</math> We cross out all choices below <math>\textbf{(C)}</math> since they are less than <math>2,</math> and <math>\textbf{(E)}</math> can't possibly be true unless either <math>\log_2{6}</math> and/or <math>\log_3{6}</math> is <math>0</math> (You can prove this by squaring.). Thus, the only feasible answer is <math>\boxed{\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}}}.</math>
  
-PureSwag
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~PureSwag
  
 
== Video Solution ==
 
== Video Solution ==
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~IceMatrix
 
~IceMatrix
  
== Video Solution ==
+
== Video Solution by OmegaLearn ==
 
https://youtu.be/RdIIEhsbZKw?t=1463
 
https://youtu.be/RdIIEhsbZKw?t=1463
  
 
~ pi_is_3.14
 
~ pi_is_3.14
  
== Video Solution (Meta-Solving Technique) ==
+
== Video Solution by OmegaLearn (Meta-Solving Technique) ==
 
https://youtu.be/GmUWIXXf_uk?t=1298
 
https://youtu.be/GmUWIXXf_uk?t=1298
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution==
 +
https://youtu.be/ObLQiTVxLco
 +
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

Revision as of 04:25, 13 November 2022

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)

Recall 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 by OmegaLearn

https://youtu.be/RdIIEhsbZKw?t=1463

~ pi_is_3.14

Video Solution by OmegaLearn (Meta-Solving Technique)

https://youtu.be/GmUWIXXf_uk?t=1298

~ pi_is_3.14

Video Solution

https://youtu.be/ObLQiTVxLco

~Education, the Study of Everything

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
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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