Question

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

• A. four solutions two of which are negative
• B. two solutions and both are negative
• C. no solution
• D. two solutions and only one of them is negative

Answer 1

Answer: (B)

$e^{4x}+8e^{3x}+13e^{2x}-8e^x+1=0$
Let $e^x=t$
Now, $t^4+8t^3+13t^2-8t+1=0$
Dividing equation by $t^2$,
$t^2+8t+13-\frac{8}{t}+\frac{1}{t^2}=0$
$t^2+\frac{1}{t^2}+8\left(t-\frac{1}{t}\right)+13=0$
${\left(t-\frac{1}{t}\right)}^2+2+8\left(t-\frac{1}{t}\right)+13=0$
Let $t-\frac{1}{t}=z$
$z^2+8z+15=0$
$(z+3)(z+5)=0$
$z=-3$ or $z=-5$
So, $t-\frac{1}{t}=-3$ or $t-\frac{1}{t}=-5$
$t^2+3t-1=0$ or $t^2+5t-1=0$
$t=\frac{-3\pm \sqrt{13}}{2}$ or $t=\frac{-5\pm \sqrt{29}}{2}$
as $t=e^x$ so $t$ must be positive,
$t=\frac{\sqrt{13}-3}{2}$ or $\frac{\sqrt{29}-5}{2}$
So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$
Hence two solution and both are negative.