The number of real solutions of the equation e^4x+4e^3x-58e^2x+4e^x+1=0 is _____ Answer: 2

Question

The number of real solutions of the equation $e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$ is \_\_\_\_\_

Answer: 2

Accepted Answer

Dividing by $e^{2x}$:
$$\left(e^x+e^{-x}\right)^2+4\left(e^x+e^{-x}\right)-60=0.$$

Let $t=e^x+e^{-x}\ge2$: $t^2+4t-60=0 \implies t=6$ (the root $t=-10$ is rejected).

$$e^x+e^{-x}=6 \implies e^{2x}-6e^x+1=0 \implies e^x=3\pm2\sqrt2,$$

both positive, giving 2 real solutions: $x=\ln(3+2\sqrt2)$ and $x=\ln(3-2\sqrt2)$.

Solutions · 1