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

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Answer 1 · 2015-06-03T02:12:55

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

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

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

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

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

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

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

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

$x = (ln(3))/2$

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