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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 9"
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The surface area is
 
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Therefore, the answer is <math>\boxed{\textbf{(E) }576}</math>.
 
Therefore, the answer is <math>\boxed{\textbf{(E) }576}</math>.
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
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==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}}
 
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Revision as of 20:40, 25 November 2021

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 1

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

Solution 2

The surface area is \begin{align*} 2 \left( \log_2 x \cdot \log_3 x + \log_2 x \cdot \log_4 x + \log_3 x \cdot \log_4 x \right) . \end{align*}

The volume is \[ \log_2 x \cdot \log_3 x \cdot \log_4 x . \]

Hence, \[ \log_2 x \cdot \log_3 x \cdot \log_4 x = 2 \left( \log_2 x \cdot \log_3 x + \log_2 x \cdot \log_4 x + \log_3 x \cdot \log_4 x \right) . \]

Dividing both sides by $\log_2 x \cdot \log_3 x \cdot \log_4 x$, we get \begin{align*} 1 & = 2 \left( \frac{1}{\log_4 x} + \frac{1}{\log_3 x} + \frac{1}{\log_2 x} \right) \\ & = 2 \left( \frac{\log_a 4}{\log_a x} + \frac{\log_a 3}{\log_a x} + \frac{\log_a 2}{\log_a x} \right) \\ & = 2 \frac{\log_a 4 + \log_a 3 + \log_a 2}{\log_a x} \\ & = 2 \frac{\log_a \left( 4 \cdot 3 \cdot 2 \right)}{\log_a x} \\ & = 2 \frac{\log_a 24}{\log_a x} \\ & = \frac{\log_a 24^2}{\log_a x} \\ & = \frac{\log_a 576}{\log_a x} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(E) }576}$. ~Steven Chen (www.professorchenedu.com)

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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