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#