Question

Как решить?              log5X2+ logx5+3=0

Answer 1

 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