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

May 5, 2016

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

#### Explanation:

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

Then the above equation becomes

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

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

or $u \left(u - 6\right) + 3 \left(u - 6\right) = 0$ or $\left(u + 3\right) \left(u - 6\right) = 0$

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

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

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

$\left({e}^{2 x} + 3\right) \ne 0$ and hence dividing above by $\left({e}^{2 x} + 3\right)$,

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

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

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