Решите пожалуйста логарифм. систему

{ log(x) [xy] = log(y) [x^2]
{ y^[2*log(y) x] = 4y- 3
=>
___ log(x) [xy] = log(x) x + log(x) y = 1 + log(x) y = 1 + 1\log(y) x
___ log(y) x^2 = 2 * log(y) x

=> первое уравнение:
1 + 1\log(y) x = 2 * log(y) x --------------> log(y) x = t
1 + 1\t = 2t -------------------------> 2t^2 - t - 1 = 0 ---------> t1 = -1\2; t2=1
t1 = -1\2 --------------> log(y) x = -1\2 -------------> x1 = y^(-1\2) = 1\y^2
t2=1 -------------------> log(y) x = 1 -----------------> x2 = y^1 = y

=> второе уравнение ------------------->y^[2*log(y) x] = 4y - 3

a) при значении ------> log(y) x = -1\2 --------------------------------------
y^[2*(-1\2)] = 4y - 3
y^(-1) = 4y-3 -----> 4y^2 - 3y - 1 = 0 ----------------------------> y1=-1\4; y(1.1) = 1
y1 = -1\4 -----------> x1 = 1\(y1)^2 = 1\(-1\4)^2 = 16 --------> x1 = 16
y(1.1) =1 -----------> x(1.1) = 1\(y1.1)^2 = 1\1^1 = 1 ---------------------> x(1,1) = 1

б) при значении ---------> log(y) x = 1 -------------------------------------------------------
y^[2*log(y) x] = 4y - 3 ------>
y^[2*1] = 4y - 3 -------> y^2 - 4y + 3 = 0 ------------------------> y2 = 1; y(2.2) = 3
y2 = 1 -------------------> x2 = 1\(y2)^2 = 1\1^1 = 1 -----------> x2 = 1
y(2.2) = 3 --------------> x(2.2) = 1\(y2.2) = 1\3 ---------------------------> x(2.2) = 1\3