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[tex]\displaystyle\bf\\1)\\\\\log_{2} 32=\log_{2} 2^{5} =5\cdot\underbrace{\log_{2} 2}_{1}=5\cdot 1=5\\\\\log_{3} 81=\log_{3} 3^{4} =4\cdot\underbrace{\log_{3} 3}_{1}=4\cdot 1=4\\\\\log_{2} \frac{1}{2} =\log_{2} 2^{-1} =-1\cdot\underbrace{\log_{2} 2}_{1}=-1\cdot 1=-1\\\\\log_{3} \frac{1}{27} =\log_{3} 3^{-3} =-3\cdot\underbrace{\log_{3} 3}_{1}=-3\cdot 1=-3[/tex]

[tex]\displaystyle\bf\\2)\\\\\log_{x} 27=3\\\\x^{3} =27\\\\x^{3} =3^{3} \\\\x=3\\\\\\\log_{\sqrt{8} } x=\frac{2}{3} \\\\x=(\sqrt{8} )^{\frac{2}{3} } =\Big[(\sqrt{8} )^{2} \Big]^{\frac{1}{3} } =8^{\frac{1}{3} } =\sqrt[3]{8} =\sqrt[3]{2^{3} } =2\\\\\\\log_{x} 0,64=-2\\\\x^{-2} =0,64\\\\x^{-2} =\Big(\frac{4}{5} \Big)^{2} \\\\x^{-2} =\Big(\frac{5}{4} \Big)^{-2} \\\\x=\frac{5}{4} =1,25[/tex]

[tex]\displaystyle\bf\\3)\\\\\log_{2} 7-\log_{2} \frac{7}{16} =\log_{2} \Big(7:\frac{7}{16} \Big)=\log_{2} \Big(7\cdot\frac{16}{7} \Big)=\log_{2} 16=\\\\=\log_{2} 2^{4} =4\cdot\log_{2} 2=4\cdot 1=4\\\\\\\log_{6} 3+\log_{6} 12=\log_{6} \Big(3\cdot 12\Big)=\log_{6} 36=\log_{6} 6^{2} =2\log_{6}6=2[/tex]