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Algebra Level 2

( 1 6 ( 2 log 10 1728 1 + 1 2 log 10 0.36 + 1 3 log 10 8 ) 1 2 ) 1 \left(\frac{1}{6}\left(\frac{2\log_{10}{1728}}{1+\frac{1}{2}\log_{10}{0.36}+\frac{1}{3}\log_{10}{8}}\right)^{\frac{1}{2}}\right)^{-1}

If the expression above can be expressed as n \sqrt{n} , find n n .


The answer is 6.

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1 solution

Chew-Seong Cheong
Feb 18, 2018

x = ( 1 6 ( 2 log 10 1728 1 + 1 2 log 10 0.36 + 1 3 log 10 8 ) 1 2 ) 1 = ( 1 6 ( 2 log 10 1728 log 10 10 + log 10 0.36 + log 10 8 3 ) 1 2 ) 1 = ( 1 6 ( 2 log 10 1728 log 10 10 + log 10 0.6 + log 10 2 ) 1 2 ) 1 = ( 1 6 ( 2 log 10 1728 log 10 ( 10 × 0.6 × 2 ) ) 1 2 ) 1 = ( 1 6 ( 2 log 10 1 2 3 log 10 12 ) 1 2 ) 1 = ( 1 6 ( 6 log 10 12 log 10 12 ) 1 2 ) 1 = ( 6 6 ) 1 = ( 1 6 ) 1 = 6 \begin{aligned} x & = \left(\frac 16\left(\frac {2\log_{10}1728}{1+\frac 12 \log_{10}0.36+\frac 13\log_{10}8}\right)^\frac 12 \right)^{-1} \\ & = \left(\frac 16\left(\frac {2\log_{10}1728}{\log_{10}10+ \log_{10}\sqrt{0.36}+\log_{10}\sqrt[3]8}\right)^\frac 12 \right)^{-1} \\ & = \left(\frac 16\left(\frac {2\log_{10}1728}{\log_{10}10+ \log_{10} 0.6 +\log_{10}2} \right)^\frac 12 \right)^{-1} \\ & = \left(\frac 16\left(\frac {2\log_{10}1728}{\log_{10}(10 \times 0.6 \times 2)} \right)^\frac 12 \right)^{-1} \\ & = \left(\frac 16\left(\frac {2\log_{10} 12^3}{\log_{10} 12} \right)^\frac 12 \right)^{-1} \\ & = \left(\frac 16\left(\frac {6 \cancel{\log_{10} 12}}{\cancel{\log_{10} 12}} \right)^\frac 12 \right)^{-1} \\ & = \left(\frac {\sqrt 6}6 \right)^{-1} = \left(\frac 1{\sqrt 6} \right)^{-1} = \sqrt 6 \end{aligned}

Therefore, n = 6 n=\boxed{6} .

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