The number of real roots of the equation e^4x-e^3x-4e^2x-e^x+1=0 is equal to Answer: 2

Question

The number of real roots of the equation $e^{4x}-e^{3x}-4e^{2x}-e^x+1=0$ is equal to

Answer: 2

Accepted Answer

Let $t=e^x>0$; dividing by $t^2$: $t^2+\dfrac1{t^2}-\left(t+\dfrac1t\right)-4=0$. With $u=t+\dfrac1t\ (\ge2)$: $u^2-2-u-4=0 \implies u^2-u-6=0 \implies (u-3)(u+2)=0 \implies u=3$.

$t+\dfrac1t=3 \implies t^2-3t+1=0$, which has 2 positive roots for $t$, giving 2 real values of $x$.

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