A^log(a)b=b
log(a)b-log(a)c=log(a)(b/c)
--------------------------------------------
{3y-x+24=27⇒x=3y-3
{log(2)[(2x-2y)/(5-y²)]=1⇒(2x-2y)/(5-y²)=2
подставим во 2
(6y-6-2y)/(5-y²)=2
4y-6=10-2y²
2y²+4y-16=0
y²+2y-8=0
y1+y2=-2 U y1*y2=-8
y1=-4⇒x1=-12-3=-15
y2=2⇒x2=6-3=3
--------------------------------
проверка
(-15;-4)
-12+15+24=27 27=27
log(2)(-30+8) не сущ
(3;2)
6-3+24=27 27=27
log(2)(6-4)-log(2)(5-4)=log(2)2-log(2)1=1-0=1 1=1
Ответ (3;2)