Respuesta :

Tsarina

[tex] \frac{ {(3)}^{3} }{ {(3)}^{y} } = \frac{1}{ {3}^{2} } \\ \\ {3}^{(3 - y)} = {3}^{ - 2} [/tex]

As bases are same,
exponents will be same too,

[tex]3 - y = - 2 \\ - y = - 5 \\ y = 5[/tex]


Therefore, y must be equal to 5.