Answer :
[tex]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}[/tex]
