# How do you solve e^(4x) + 4e^(2x) + -21 = 0 ?

Jun 3, 2015

Let's let $u = {e}^{2 x}$

Substituting it into the equation, we get
${u}^{2} + 4 u - 21 = 0$

We can factor this into:
$\left(u + 7\right) \left(u - 3\right) = 0$

Substituting ${e}^{2 x}$ back in:
$\left({e}^{2 x} + 7\right) = 0 \mathmr{and} \left({e}^{2 x} - 3\right) = 0$

${e}^{2 x} = - 7 \mathmr{and} {e}^{2 x} = 3$

$2 x = \ln \left(- 7\right) \mathmr{and} 2 x = \ln \left(3\right)$

$x = \frac{\ln \left(- 7\right)}{2} \mathmr{and} x = \frac{\ln \left(3\right)}{2}$

However, a negative argument in a logarithm will not produce a real solution. Therefore, the only real solution is

$x = \frac{\ln \left(3\right)}{2}$