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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 9"
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<math>\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576</math>
 
<math>\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576</math>
  
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==Solution 1==
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==Solution==
 
The surface area of this right rectangular prism is <math>2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x).</math>
 
The surface area of this right rectangular prism is <math>2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x).</math>
  
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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== Solution 3 ==
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==Video Solution by TheBeautyofMath==
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The surface area is
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https://youtu.be/wlDlByKI7A8?t=649
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<cmath>
 
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\begin{align*}
 
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2 \left( \log_2 x \cdot \log_3 x + \log_2 x \cdot \log_4 x + \log_3 x \cdot \log_4 x \right) .
 
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\end{align*}
 
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</cmath>
 
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The volume is
 
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<cmath>
 
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\[
 
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\log_2 x \cdot \log_3 x \cdot \log_4 x .
 
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\]
 
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</cmath>
 
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Hence,
 
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<cmath>
 
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\[
 
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\log_2 x \cdot \log_3 x \cdot \log_4 x
 
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= 2 \left( \log_2 x \cdot \log_3 x + \log_2 x \cdot \log_4 x + \log_3 x \cdot \log_4 x \right)
 
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.
 
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\]
 
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</cmath>
 
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Dividing both sides by <math>\log_2 x \cdot \log_3 x \cdot \log_4 x</math>, we get
 
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<cmath>
 
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\begin{align*}
 
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1 & = 2 \left( \frac{1}{\log_4 x} + \frac{1}{\log_3 x} + \frac{1}{\log_2 x} \right) \\
 
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& = 2 \left( \frac{\log_a 4}{\log_a x} + \frac{\log_a 3}{\log_a x} + \frac{\log_a 2}{\log_a x} \right) \\
 
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& = 2 \frac{\log_a 4 + \log_a 3 + \log_a 2}{\log_a x} \\
 
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& = 2 \frac{\log_a \left( 4 \cdot 3 \cdot 2 \right)}{\log_a x} \\
 
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& = 2 \frac{\log_a 24}{\log_a x} \\
 
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& = \frac{\log_a 24^2}{\log_a x} \\
 
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& = \frac{\log_a 576}{\log_a x} .
 
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\end{align*}
 
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</cmath>
 
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Therefore, the answer is <math>\boxed{\textbf{(E) }576}</math>.
 
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~Steven Chen (www.professorchenedu.com)
 
  
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~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:58, 7 April 2022

Problem

A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_{2}x, \log_{3}x,$ and $\log_{4}x.$ What is $x?$

$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576$

Solution

The surface area of this right rectangular prism is $2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x).$

The volume of this right rectangular prism is $\log_{2}x\log_{3}x\log_{4}x.$

Equating the numerical values of the surface area and the volume, we have \[2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x)=\log_{2}x\log_{3}x\log_{4}x.\] Dividing both sides by $\log_{2}x\log_{3}x\log_{4}x,$ we get \[2\left(\frac{1}{\log_{4}x}+\frac{1}{\log_{3}x}+\frac{1}{\log_{2}x}\right)=1. \hspace{15mm} (\bigstar)\] Recall that $\log_{b}a=\frac{1}{\log_{a}b}$ and $\log_{b}\left(a^n\right)=n\log_{b}a,$ so we rewrite $(\bigstar)$ as \begin{align*} 2(\log_{x}4+\log_{x}3+\log_{x}2)&=1 \\ 2\log_{x}24&=1 \\ \log_{x}576&=1 \\ x&=\boxed{\textbf{(E)}\ 576}. \end{align*} ~MRENTHUSIASM

Video Solution by TheBeautyofMath

https://youtu.be/wlDlByKI7A8?t=649

~IceMatrix

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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