We have, e4x−e3x−4e2x−ex+1=0
Let ex=t, we get
t4−t3−4t2−t+1=0
Divide with t2 on both sides
⇒t2−t−4−t1+t21=0
⇒(t+t1)2−(t+t1)−6=0
Let α=t+t1≥2, we get
α2−α−6=0
⇒(α−3)(α+2)=0
⇒α=3,−2 (reject)
⇒t+t1=3⇒t2−3t+1=0
⇒ex=23±5
⇒ex=23±2.23(∵5≈2.23)
⇒ex=25.23 or ex=20.77
⇒ex=2.615 or ex=0.385

⇒ The number of real roots =2.