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Lemma
Exponentials and Logarithms Difficulty 4

TMUA practice: Exponentials Logs, problem 1

A TMUA-calibre exponentials logs problem at difficulty 4 of 5, with the full worked solution.

The question
A right rectangular prism whose surface area and volume are numerically equal has edge lengths log⁡2x,log⁡3x,\log_{2}x, \log_{3}x, and log⁡4x.\log_{4}x. What is x?x?
A 262\sqrt{6}
B 666\sqrt{6}
C 2424
D 4848
E 576576

The correct answer is highlighted. E

Worked solution

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

Equating the numerical values of the surface area and the volume, we have

2(log⁡2xlog⁡3x+log⁡2xlog⁡4x+log⁡3xlog⁡4x)=log⁡2xlog⁡3xlog⁡4x.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⁡2xlog⁡3xlog⁡4x,\log_{2}x\log_{3}x\log_{4}x, we get

2(1log⁡4x+1log⁡3x+1log⁡2x)=1.(★)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⁡ba=1log⁡ab\log_{b}a=\frac{1}{\log_{a}b} and log⁡b(an)=nlog⁡ba,\log_{b}\left(a^n\right)=n\log_{b}a, so we rewrite (★)(\bigstar) as

2(log⁡x4+log⁡x3+log⁡x2)=12log⁡x24=1log⁡x576=1x=(E) 576.\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*}