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

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Answer 1 · 2015-06-03T02:12:55

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

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

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

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

$e^{2 x} = - 7 and e^{2 x} = 3$ $e^(2x) = -7 and e^(2x) =3$

$2 x = ln (- 7) and 2 x = ln (3)$ $2x = ln(-7) and 2x = ln(3)$

$x = \frac{ln (- 7)}{2} and x = \frac{ln (3)}{2}$ $x = (ln(-7))/2 and x = (ln(3))/2$

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

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

How do you solve e^(4x) + 4e^(2x) + -21 = 0 e4x+4e2x+−21=0e^(4x) + 4e^(2x) + -21 = 0 ?