Question

2 log^2x+log x-1=0 como resolver?

Answer 1

2log²ˣ + logx - 1 = 0 , C.E: x > 0

2(logˣ)² + logx - 1 = 0

Seja logx = y

2y² + y - 1 = 0

Δ = 1² - 4.2(-1)

Δ = 1 + 9

Δ = 9

y = (-1 - 3)/4 = -1 ou 

y = ( -1 + 3)/2 = 1

Então

logx = -1 => x = 10⁻¹ => x = 1/10

 ou 

logx = 1 => x = 10¹ => x = 10