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Algebra Difficulty 5.0 AIME, harder Find the answer United States

A right rectangular prism whose surface area and volume are numerically equal has edge lengths log2x\log_2 x, log3x\log_3 x, and log4x\log_4 x. What is xx?

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Solution

Solution:
Denote the edge lengths by a=log2xa = \log_2 x, b=log3xb = \log_3 x, and c=log4xc = \log_4 x. The condition that the surface area numerically equals the volume is equivalent to 2(ab+ac+bc)=abc2(ab + ac + bc) = abc. Dividing both sides by 2abc2abc gives 1a+1b+1c=12\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}. By the Change of Base Formula, 1a=log2x\frac{1}{a} = \log_2 x, 1b=log3x\frac{1}{b} = \log_3 x, and 1c=log4x\frac{1}{c} = \log_4 x. Therefore log22+log33+log44=12\log_2 2 + \log_3 3 + \log_4 4 = \frac{1}{2}, so log2(234)=12\log_2(2 \cdot 3 \cdot 4) = \frac{1}{2}. From x12=24x^{\frac{1}{2}} = 24 it follows that x=242=576x = 24^2 = 576.

Because the surface area and volume are numerically equal,
2(log2xlog3x+log2xlog4x+log3xlog4x)=log2xlog3xlog4x. 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.
By the Change of Base Formula with base 10,
2(logxlog2logxlog3+logxlog2logxlog4+logxlog3logxlog4)=logxlog2logxlog3logxlog4. 2 \left( \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 3} + \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 4} + \frac{\log x}{\log 3} \cdot \frac{\log x}{\log 4} \right) = \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 3} \cdot \frac{\log x}{\log 4}.
Multiplying both sides by log2log3log4\log 2 \cdot \log 3 \cdot \log 4 and simplifying gives
2(logx)2(log4+log3+log2)=(logx)3. 2 (\log x)^2 (\log 4 + \log 3 + \log 2) = (\log x)^3.
Using log properties and moving all terms to one side gives
(logx)2(2log(432)logx)=0. (\log x)^2 (2 \log (4 \cdot 3 \cdot 2) - \log x) = 0.
Thus either logx=0\log x = 0 or 2log(432)logx=02\log(4 \cdot 3 \cdot 2) - \log x = 0. If logx=0\log x = 0 then x=1x = 1, but this would make each edge length of the prism 0, which is impossible. Therefore 2log(432)logx=02\log(4 \cdot 3 \cdot 2) - \log x = 0. This means logx=log(242)\log x = \log (24^2) and it follows that x=242=576x = 24^2 = 576.

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