Answer:
z = 1 + 9y³
Step-by-step explanation:
It looks like you have ...
[tex]\log_3{(z-1)}-\log_z{(z)}=1+3\log_3{(y)}\\\\\log_3{(z-1)}-1=1+\log_3{(y^3)}\\\\\log_3{(z-1)}=2+\log_3{(y^3)}\qquad\text{add 1}\\\\z-1=(3^2)(y^3)\qquad\text{take antilog}\\\\\boxed{z=1+9y^3}\qquad\text{add 1}[/tex]
