Difference between revisions of "2021 Fall AMC 12A Problems/Problem 9"

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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Video Solution by TheBeautyofMath==
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https://youtu.be/wlDlByKI7A8?t=649
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~IceMatrix
  
 
==See Also==
 
==See Also==
 
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{{AMC12 box|year=2021 Fall|ab=A|num-b=8|num-a=10}}
 
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{{MAA Notice}}

Latest revision as of 20:58, 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 (ProblemsAnswer KeyResources)
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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