The sum of all real roots of equation e 2 x - 4 6 e 2 x - 5 e x + 1 = 0 is

Step-by-step solution

Let e x = t so, e 2 x = t 2 So, the given equation becomes, t 2 - 4 6 t 2 - 5 t + 1 = 0 t 2 - 4 = 0 (or) 6 t 2 - 5 t + 1 = 0 t 2 = 4 (or) 6 t 2 - 3 t - 2 t + 1 = 0 t = ± 2 (or) 3 t 2   t - 1 - 1 2   t - 1 = 0 t = ± 2 (or) 3 t - 1 2 t - 1 = 0 t = ± 2 (or) t = 1 3 (or) t = 1 2 But, e x = t > 0    ∵     e x > 0 ; ∀ x ∈ R ∴ Only possible values for t are 2 , 1 3 , 1 2 When t = 2 ,     e x = 2   ⇒ x = log e 2 (or) ln 2 When