What is $y$ in $$3^{2y}\cdot3^{\log_{3}(1/3)}=9$$
I apologize if this is confusing, i wasn't sure how to type this equation in here to ask it. If you can please show the steps it would help me because I have a test coming soon. Any help is appreciated!
$$\log_3(1/3)=\log_33^{-1}=-1$$ Thus, $$3^{2y}\times 3^{\log_3(1/3)}=3^{2y}\times 3^{-1}=3^{2y-1}=9=3^2\\ \implies 2y-1=2\\ \implies y=\dfrac{3}{2}.$$