How do you solve $e^{4 x} - 3 e^{2 x} - 18 = 0$ $e^(4x)-3e^(2x)-18=0$ ?

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Answer 1 · 2016-05-05T16:38:17

$x = \frac{1}{2} ln 6$ $x=1/2ln6$

To solve $e^{4 x} - 3 e^{2 x} - 18 = 0$ $e^(4x)-3e^(2x)-18=0$ , let $e^{2 x} = u$ $e^(2x)=u$

Then the above equation becomes

$u^{2} - 3 u - 18 = 0$ $u^2-3u-18=0$ and splitting middle term we get

$u^{2} - 6 u + 3 u - 18 = 0$ $u^2-6u+3u-18=0$

or $u (u - 6) + 3 (u - 6) = 0$ $u(u-6)+3(u-6)=0$ or $(u + 3) (u - 6) = 0$ $(u+3)(u-6)=0$

substituting $u = e^{2 x}$ $u=e^(2x)$ , we get

$(e^{2 x} + 3) (e^{2 x} - 6) = 0$ $(e^(2x)+3)(e^(2x)-6)=0$

As $e^{2 x}$ $e^(2x)$ is always positive

$(e^{2 x} + 3) \neq 0$ $(e^(2x)+3)!=0$ and hence dividing above by $(e^{2 x} + 3)$ $(e^(2x)+3)$ ,

$(e^{2 x} - 6) = 0$ $(e^(2x)-6)=0$

i.e. $e^{2 x} = 6$ $e^(2x)=6$

or $2 x = ln 6$ $2x=ln6$ and hence $x = \frac{1}{2} ln 6$ $x=1/2ln6$

How do you solve e4x−3e2x−18=0e^(4x)-3e^(2x)-18=0?