e^4x + e^3x – 4e^2x + e^x + 1 = 0

Dividing by e^2x, we get

e^2x + e^x - 4 + $${1 \over {{e^x}}}$$ + $${1 \over {{e^{2x}}}}$$ = 0

$$ \Rightarrow $$ $$\left( {{e^{2x}} + {1 \over {{e^{2x}}}}} \right) + \left( {{e^x} + {1 \over {{e^x}}}} \right)$$ - 4 = 0

$$ \Rightarrow $$ $${\left( {{e^x} + {1 \over {{e^x}}}} \right)^2} - 2$$ + $$\left( {{e^x} + {1 \over {{e^x}}}} \right)$$ - 4 = 0

Let $${{e^x} + {1 \over {{e^x}}} = z}$$

(z² – 2) + (z) – 4 = 0

$$ \Rightarrow $$ z² + z – 6 = 0

$$ \Rightarrow $$ z = –3, 2

$$ \therefore $$ $${{e^x} + {1 \over {{e^x}}} = 2}$$

$$ \Rightarrow $$ (e^x – 1)² = 0 $$ \Rightarrow $$ x = 0.

$$ \therefore $$ Number of real roots = 1