Given equation:
\[
e^{4x}-e^{3x}-4e^{2x}-e^x+1=0
\]
Let
\[
y=e^x \quad (y > 0)
\]
Substituting,
\[
y^4-y^3-4y^2-y+1=0
\]
This is a reciprocal equation of the form:
\[
ay^4+by^3+cy^2+by+a=0
\]
Since \( y \neq 0 \), divide throughout by \( y^2 \):
\[
y^2-y-4-\frac{1}{y}+\frac{1}{y^2}=0
\]
Rearranging,
\[
\left(y^2+\frac{1}{y^2}\right)-\left(y+\frac{1}{y}\right)-4=0
\]
Let
\[
z=y+\frac{1}{y}
\Rightarrow y^2+\frac{1}{y^2}=z^2-2
\]
Substitute:
\[
(z^2-2)-z-4=0
\Rightarrow z^2-z-6=0
\]
\[
(z-3)(z+2)=0
\Rightarrow z=3 \text{ or } z=-2
\]
Since \( y > 0 \), by AM–GM:
\[
y+\frac{1}{y} \ge 2
\]
Hence \( z=-2 \) is rejected.
So,
\[
y+\frac{1}{y}=3
\]
Multiplying by \( y \):
\[
y^2-3y+1=0
\]
\[
y=\frac{3\pm\sqrt5}{2}
\]
Both roots are positive, hence both are valid.
Since \( y=e^x \), each value of \( y \) gives one real value of \( x \).
\[
\boxed{\text{Number of real roots }=2}
\]
