# Can you help me find the answer?

## ${e}^{4 x} + 4 {e}^{2 x} - 21 = 0$

Apr 13, 2017

$x = 1.0986$

#### Explanation:

In ${e}^{4 x} + 4 {e}^{2 x} - 21 = 0$, let ${e}^{2 x} = u$

then the equation becomes

${u}^{2} + 4 u - 21 = 0$

or ${u}^{2} + 7 u - 3 u - 21 = 0$

or $u \left(u + 7\right) - 3 \left(u + 7\right) = 0$

or $\left(u - 3\right) \left(u + 7\right) = 0$

bur as $u = {e}^{x}$, $u + 7 \ne 0$ and dividing by $u + 7$, we get

$u - 3 = 0$ or ${e}^{x} = 3$ i.e. $x = \ln 3 = 1.0986$