Как решить? log5X2 + logx5+3=0

log5 x^2 + logx 5 + 3 = 0

ОДЗ: x>0

2 * log5 x + logx 5 + 3 = 0

2 * (1/logx 5) + logx 5 + 3 = 0

2 + (logx 5) ^2 + 3*logx 5 = 0

(logx 5) ^2 + 3*logx 5 + 2 = 0

logx 5 = t

t^2 + 3t + 2 = 0

t1 = - 2 = > logx 5 = - 2 = > x^ (-2) = 5 = > 1/x^2=5 = > x=+-1/V5=+-V5/5, по ОДЗ x=+V5/5

t2 = - 1 = > logx 5 = - 1 = > x^ (-1) = 5 = > 1/x=5 = > x=1/5

Ответ: x=+V5/5 или x=1/5