Question

Помогите пожалуйста)

Answer 1

несколько теории
b^log(a) c = c^log(a) b
log (a^n) b^m = m/n log(a) b
log(a) x + log(a) y = log(a) xy
ОДЗ log(x) y     x>0 x≠1 y>0
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ОДЗ
x>0 y>0
x^log(3) y = y^log(3) x
20x^log(3)y + 7y^log(3)x = 81*³√3
20y^log(3)x + 7y^log(3)x = 81*³√3
27y^log(3)x = 81*³√3
y^log(3)x = 3*³√3 = 3^(4/3)

log(3^2) x^2 + log(3^3) y^3 = 9/3
log(3) x + log(3) y = 3
log(3) xy = 3
xy=27
x=27/y

y^log(3) 27/y = 3^4/3
логарифмируем по основанию 3
log(3) 27/y * log(3) y = log(3) 3^4/3
log(3) y * (3 - log(3) y) = 4/3
log(3) y = t
3t - t^2 = 4/3
t^2 - 3t + 4/3=0
D=b^2-4ac = 9 - 16/3 = (27-16)/3 = 11/3
t12= (3+-√(11/3))/2
y12 = 3^(3+-√(11/3))/2
x12 =  27 / 3^(3+-√(11/3))/2
ответ (27 / 3^(3+√(11/3))/2  3^(3+√(11/3))/2)  (27 / 3^(3-√(11/3))/2  3^(3-√(11/3))/2