Given,
e8x−e6x−3e4x−e2x+1=0
⇒e4x−e2x−3−e2x1+e4x1=0
⇒e4x+e4x1+2−(e2x+e2x1)=5
⇒(e2x+e2x1)2−(e2x+e2x1)=3
Now let (e2x+e2x1)=t
So, the equation becomes
t2−t−5=0
⇒t=21+21, ignoring negative sign as exponential function are positive,
Now e2x+e2x1=21+21
⇒e4x−21+21e2x+1=0 which is quadratic equation in e2x with upward parabola,
Now by A.M≥G.M we get, e2x+e2x1≥2
So, y=21+21>2, hence it will cut at two distinct point,
Hence, there will be two solution.