The equation e 4 x + 8 e 3 x + 13 e 2 x - 8 e x + 1 = 0 , x ? R has [2023]

Question

The equation $e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0, x \in R$ has [2023]

Smart Achievers · Accepted Answer

(2)

$Given, e^{4 x} + 8 e^{3 x} + 13 e^{2 x} - 8 e^{x} + 1 = 0$

$Let e^{x} = t$

Now, $t^{4} + 8 t^{3} + 13 t^{2} - 8 t + 1 = 0 ...(i)$

$Dividing equation (i) by t^{2}, we get$

$\Rightarrow t^{2} + 8 t + 13 - \frac{8}{t} + \frac{1}{t^{2}} = 0 \Rightarrow t^{2} + \frac{1}{t^{2}} + 8 (t - \frac{1}{t}) + 13 = 0$

$\Rightarrow {(t - \frac{1}{t})}^{2} + 2 + 8 (t - \frac{1}{t}) + 13 = 0$

$Let t - \frac{1}{t} = z$

$\Rightarrow z^{2} + 8 z + 15 = 0 \Rightarrow (z + 3) (z + 5) = 0$

$\Rightarrow z = - 3 or z = - 5$ $So, t - \frac{1}{t} = - 3 or t - \frac{1}{t} = - 5$

$\Rightarrow t^{2} + 3 t - 1 = 0 or t^{2} + 5 t - 1 = 0;$ $\Rightarrow t = \frac{- 3 \pm \sqrt{13}}{2} or t = \frac{- 5 \pm \sqrt{29}}{2}$

$As t = e^{x}, it must be positive.$ $\Rightarrow t = \frac{\sqrt{13} - 3}{2} or t = \frac{\sqrt{29} - 5}{2}$

$\Rightarrow x = \ln (\frac{\sqrt{13} - 3}{2}) or x = \ln (\frac{\sqrt{29} - 5}{2})$

$Hence, two solutions are possible and both are negative.$

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