tolong nomor 7&8 ya terima kasih

[tex] \frac{( \sqrt[3]{ \frac{1}{243} } )^{3x}}{3^{x-4}} =1 \\ \\ \frac{3^{-5}^x}{3^{x-4}}= 1 \\ \\ 3^{-6x+4}=3^0 \\ \\ -6x+4=0 \\ -6x=-4 \\ x= \frac{-4}{-6} = \frac{ 2}{3}[/tex]