Verified answer

Смотри........................

[tex]\displaystyle\bf\\\frac{1}{2} \log_{3} \frac{4}{81} -\frac{1}{3} \log_{3} \frac{8}{27} =\log_{3} \Big(\frac{4}{81} \Big)^{\dfrac{1}{2} } -\log_{3} \Big(\frac{8}{27}\Big)^{\dfrac{1}{3} } =\\\\\\=\log_{3} \sqrt{\frac{4}{81} } -\log_{3} \sqrt[3]{\frac{8}{27} } =\log_{3} \sqrt{\Big(\frac{2}{9}\Big)^{2} } -\log_{3} \sqrt[3]{\Big(\frac{2}{3} \Big)^{3} } =[/tex]

[tex]\displaystyle\bf\\=\log_{3} \frac{2}{9} -\log_{3} \frac{2}{3}=\log_{3} \Big(\frac{2}{9} :\frac{2}{3} \Big)=\log_{3} \Big(\frac{2}{9} \cdot\frac{3}{2} \Big)=\log_{3} \frac{1}{3} =\\\\\\=\log_{3} 3^{-1} =-1\cdot\underbrace{\log_{3} 3}_{1}=-1\cdot 1=-1[/tex]