Write as a single logarithm 3-1/2(log6+log3-3log2)

\( 3-\dfrac{1}{2}\left(\log6+\log3-3\log2\right)=\\ \log10^3-\dfrac{1}{2}\left(\log\left(6\cdot3\right)-\log2^3\right)=\\ \log1000-\dfrac{1}{2}\left(\log18-\log8\right)=\\ \log1000-\dfrac{1}{2}\left(\log\dfrac{18}{8}\right)=\\ \log1000-\dfrac{1}{2}\left(\log\dfrac{9}{4}\right)=\\ \log1000-\log\left(\dfrac{9}{4}\right)^{\dfrac{1}{2}}=\\ \log1000-\log\sqrt{\dfrac{9}{4}}=\\ \log1000-\log\dfrac{3}{2}=\\ \log\dfrac{1000}{\frac{3}{2}}=\\ \log\left(1000\cdot\dfrac{2}{3}\right)=\\ \log\dfrac{2000}{3} \)

Answer in document below.

\( 3-\dfrac{1}{2}\left(\log6+\log3-3\log2\right)=\\ \log10^3-\dfrac{1}{2}\left(\log\left(6\cdot3\right)-\log2^3\right)=\\ \log1000-\dfrac{1}{2}\left(\log18-\log8\right)=\\ \log1000-\dfrac{1}{2}\left(\log\dfrac{18}{8}\right)=\\ \log1000-\dfrac{1}{2}\left(\log\dfrac{9}{4}\right)=\\ \log1000-\log\left(\dfrac{9}{4}\right)^{\dfrac{1}{2}}=\\ \log1000-\log\sqrt{\dfrac{9}{4}}=\\ \log1000-\log\dfrac{3}{2}=\\ \log\dfrac{1000}{\frac{3}{2}}=\\ \log\left(1000\cdot\dfrac{2}{3}\right)=\\ \log\dfrac{2000}{3} \)  Answer in document below.

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