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

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Answer 1 · 2016-05-05T16:38:17

$x=1/2ln6$

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

Then the above equation becomes

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

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

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

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

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

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

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

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

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

or $2x=ln6$ and hence $x=1/2ln6$

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