The equation e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0$

Let $e^x=t$

Now, $t^4+8 t^5+13 t^2-8 t+1=0$

Dividing equation by $t ^2$,

$t^2+8 t+13-\frac{8}{t}+

Vedclass · Accepted Answer

two solutions and both are negative

Vedclass · Answer

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

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

Similar Questions

If $x$ is real, then the value of $\frac{{{x^2} + 34x - 71}}{{{x^2} + 2x - 7}}$ does not lie between

Let $x$ and $y$ be two $2-$digit numbers such that $y$ is obtained by reversing the digits of $x$. Suppose they also satisfy $x^2-y^2=m^2$ for some positive integer $m$. The value of $x+y+m$ is

If $x$ be real, the least value of ${x^2} - 6x + 10$ is

Suppose that $x$ and $y$ are positive number with $xy = \frac{1}{9};\,x\left( {y + 1} \right) = \frac{7}{9};\,y\left( {x + 1} \right) = \frac{5}{{18}}$ . The value of $(x + 1) (y + 1)$ equals

Let $a, b, c$ be non-zero real numbers such that $a+b+c=0$, let $q=a^2+b^2+c^2$ and $r=a^4+b^4+c^4$. Then,