a, `3^2 . 3^(3.n) = 3^5`
`⇔ 2 + ( 3 . n ) = 5`
`⇔ 3 . n = 3`
`⇔ n = 1`
Vậy `n = 1`
b, `( 2^3 : 4 ) . 2^n = 4`
`⇔ ( 2^3 : 2^2 ) . 2^n = 2^2`
`⇔ ( 3 - 1 ) . n = 2`
`⇔ 2 . n = 2`
`⇔ n = 1`
Vậy `n = 1`
c, `3^(-2) . 3^4 . 3^n = 3^7`
`⇔ -2 + 4 + n = 7`
`⇔ 2 + n = 7`
`⇔ n = 5`
Vậy `n = 5`
d, `2^(-1) . 2^n + 4 . 2^n = 9 . 2^5`
`⇔ 2^(-1) + 2^n + 2^2 . 2^n = 9 . 2^5`
`⇔ 2^n . 9/2 = 9 . 2^5`
`⇔ 2^n . 1 = 2^6`
`⇔ n = 6`
Vậy `n = 6`