# How do you solve e^(2x)+8e^x+12 = 0 ?

Jun 21, 2015

This has no real solutions, but if you allow complex values it has solutions:

$x = {\log}_{e} 2 + i \left(2 n + 1\right) \pi$ for all $n \in \mathbb{Z}$

and

$x = {\log}_{e} 6 + i \left(2 n + 1\right) \pi$ for all $n \in \mathbb{Z}$

#### Explanation:

$0 = {e}^{2 x} + 8 {e}^{x} + 12 = \left({e}^{x} + 6\right) \left({e}^{x} + 2\right)$

So we are looking for $x$ such that ${e}^{x} = - 2$ or ${e}^{x} = - 6$.

There are no real values of $x$ that satisfy this, but we can use ${e}^{i \pi} = - 1$ and the properties of exponents to find solutions:

$x = {\log}_{e} 2 + i \left(2 n + 1\right) \pi$ for all $n \in \mathbb{Z}$

and

$x = {\log}_{e} 6 + i \left(2 n + 1\right) \pi$ for all $n \in \mathbb{Z}$

For example,

${e}^{{\log}_{e} 2 + i \left(2 n + 1\right) \pi}$

$= {e}^{{\log}_{e} 2} \cdot {\left({e}^{i \pi}\right)}^{2 n + 1}$

$= 2 \cdot {\left(- 1\right)}^{2 n + 1}$

$= - 2$