Помогите с решением систем уравнений {log (3) (x-2) = 25*5^y

{ 5^-y+log (3) (x-2) = 10

И {log (2) (x+1) = 64*2^y

{2^-y+log (2) (x+1) = 16

1) 10-5^-y=25*5^y

10*5^y-1=25*5^2y

5^y=z

25z^2-10z+1=0

(5z-1) ^2=0

z=5^-1

y=-1

lo (3) (x-2) = 5

x-2=243

x=245

Ответ: (245,-1)

2)

log (2) (x+1) = 16-2^y

16-2^-y=64*2^y

2^y=z

16z-1=64z^2

(8z-1) ^2=0

z=2^-3

y=-3

log (2) (x+1) = 8

x+1=256

x=255

Ответ: (255,-3)