Cześć!
Od a) do d)
[tex]\log_\frac{1}{3}\frac{2}{3}+\log_\frac{1}{3}\frac{1}{6}=\log_\frac{1}{3}(\frac{2}{3}\cdot\frac{1}{6})=\log_\frac{1}{3}\frac{1}{9}=\log_\frac{1}{3}(\frac{1}{3})^2=2\\\\\log_2(64\cdot128)=\log_264+\log_2128=\log_22^6+\log_22^7=6+7=13\\\\2\log_39=2\log_33^2=2\cdot2=4\\\\\log_\frac{1}{9}(\log_28)=\log_\frac{1}{9}(\log_22^3)=\log_\frac{1}{9}3=\log_\frac{1}{9}(\frac{1}{9})^{-\frac{1}{2}}=-\frac{1}{2}[/tex]
Od e) do h)
[tex]\log8+\log125=\log(8\cdot125)=\log1000=\log10^3=3\\\\\log_515-\log_575=\log_5\frac{15}{75}=\log_5\frac{1}{5}=\log_55^{-1}=-1\\\\\log_3(27\cdot81)=\log_327+\log_381=\log_33^3+\log_33^4=3+4=7\\\\3\log_24=3\log_22^2=3\cdot2=6[/tex]
