Question

Hello, how to do 12(c) ?:)

Answer 1

(
`(4)/(11)` ,
`(1)/(11)` )

Explanation:

27 = 3³ and √9 = 3 and xy ≠ 0

`(3^(3x) )/(3^(y) )` = 3 ⇔
`3^(3x-y)` =
`3^(1)` ⇒ 3x - y = 1

`log_(2) x` - 2 =
`log_(2) y` ⇔
`log_(2) x` -
`log_(2) y` = 2 ⇔
`log_(2) (x)/(y)` = 2 ⇒
`(x)/(y)` = 2² ⇒ x = 4y

`\left \{ {{3x-y=1} \atop {x=4y}} \right.`

3(4y) - y = 1 ⇒ y =
`(1)/(11)` and x =
`(4)/(11)`

(
`(4)/(11)` ,
`(1)/(11)` )

Answer 2

Explanation:

27^x / sqrt(9^y) = 3 and log2(x) - 2 = log2(y).

3^3x / 3^y = 3 and log2(x) - log2(4) = log2(y).

3x - y = 1 and x/4 = y.

Therefore x = 4/11 and y = 1/11.