Given, \(\bigg(3^{\frac12}\bigg)^5 \times(3^2)^2 =3^n\times3\times3^{\frac12}\)⇒ \(3^{\frac52}\times3^4=3^{n+1+\frac12}\)
⇒ \(3^{\frac52+4}=3^{n+\frac32}\) ⇒ \(3^{\frac{13}{2}}=3^{n+\frac32}\) ⇒ n+\(\frac32=\frac{13}{2}\) ⇒ n = \(\frac{13}{2}-\frac{3}{2} = \frac{10}2 = 5.\)