ОДЗ: x>0, y>0
Log5(x)+y=7
x^y= 5^12
log5(x) = 7-y
x= 5^(7-y)
(5^(7-y))^y= 5^12
7y-y^2=12
y^2-7y+12=0
D= 49-48=1
y1= (7+1)/2= 4
y2= (7-1)/2= 3
x1= 5^3= 125
x2= 5^4= 625
ответ; (125;4), (625;3)
