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

Step-by-step 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.
Answer:
( [tex]\frac{4}{11}[/tex] , [tex]\frac{1}{11}[/tex] )
Step-by-step explanation:
27 = 3³ and √9 = 3 and xy ≠ 0
[tex]\frac{3^{3x} }{3^{y} }[/tex] = 3 ⇔ [tex]3^{3x-y}[/tex] = [tex]3^{1}[/tex] ⇒ 3x - y = 1
[tex]log_{2} x[/tex] - 2 = [tex]log_{2} y[/tex] ⇔ [tex]log_{2} x[/tex] - [tex]log_{2} y[/tex] = 2 ⇔ [tex]log_{2} \frac{x}{y}[/tex] = 2 ⇒ [tex]\frac{x}{y}[/tex] = 2² ⇒ x = 4y
[tex]\left \{ {{3x-y=1} \atop {x=4y}} \right.[/tex]
3(4y) - y = 1 ⇒ y = [tex]\frac{1}{11}[/tex] and x = [tex]\frac{4}{11}[/tex]
( [tex]\frac{4}{11}[/tex] , [tex]\frac{1}{11}[/tex] )